Moving Charges and Magnetic Field — Competition Zone

A proton and an alpha particle both enter a region of uniform magnetic field B, moving at right angles to the field B. If the radius of circular orbits for both the particles is equal and the kinetic energy acquired by proton is 1 MeV, the energy acquired by the alpha particle will be :

1. 0.5 MeV
2. 1.5 MeV
3. 1 MeV
4. 4 MeV

Answer

1 MeV

Reason — Given, the radii of the circular orbits are equal and K_p = 1 MeV.

For a particle entering perpendicular to the field,

$\text r = \dfrac{\sqrt{2\text{mK}}}{\text{qB}}$

As the radii are equal,

$\dfrac{\sqrt{2\text m_\text p \text K_\text p}}{\text e} = \dfrac{\sqrt{2\text m_α \text K_α}}{2\text e}$

Putting m_α = 4m_p,

$\sqrt{\text m_\text p \text K_\text p} = \dfrac{\sqrt{4\text m_\text p \text K_α}}{2} \\[1em] \sqrt{\text m_\text p \text K_\text p} = \sqrt{\text m_\text p \text K_α} \\[1em] \text K_α = \text K_\text p = 1\ \text{MeV}$

An electron moving in a circular orbit of radius r makes n revolutions per second. The magnetic field produced at the centre has magnitude :

1. zero
2. $\dfrac{\text µ_0 \text n^2 \text e}{\text r}$
3. $\dfrac{\text µ_0 \text{ne}}{2\text r}$
4. $\dfrac{\text µ_0 \text{ne}}{2\text{πr}}$

Answer

$\dfrac{\text µ_0 \text{ne}}{2\text r}$

Reason — Given, the electron makes n revolutions per second in a circle of radius r.

The revolving electron is equivalent to a current loop, and the equivalent current is

$\text I = \dfrac{\text e}{\text T} \\[1em] \text I = \text{ne}$

The magnetic field at the centre of a circular loop is

$\text B = \dfrac{\text µ_0 \text I}{2\text r} \\[1em] \text B = \dfrac{\text µ_0 \text{ne}}{2\text r}$

A wire carrying a current I has the shape as shown in figure. Linear parts of the wire are very long and parallel to the X-axis while semi-circular portion of radius R is lying in Y-Z plane. Magnetic field at point O is :

1. $\vec{\text B} = -\dfrac{\text µ_0}{4\text π}\dfrac{\text I}{\text R}(\text π\hat{\text i} - 2\hat{\text k})$
2. $\vec{\text B} = -\dfrac{\text µ_0}{4\text π}\dfrac{\text I}{\text R}(\text π\hat{\text i} + 2\hat{\text k})$
3. $\vec{\text B} = \dfrac{\text µ_0}{4\text π}\dfrac{\text I}{\text R}(\text π\hat{\text i} - 2\hat{\text k})$
4. $\vec{\text B} = \dfrac{\text µ_0}{4\text π}\dfrac{\text I}{\text R}(\text π\hat{\text i} + 2\hat{\text k})$

Answer

$\vec{\text B} = -\dfrac{\text µ_0}{4\text π}\dfrac{\text I}{\text R}(\text π\hat{\text i} + 2\hat{\text k})$

Reason — Given, two very long linear parts parallel to the X-axis and a semi-circular portion of radius R lying in the Y-Z plane.

Field due to the linear parts : Each linear part is a semi-infinite conductor whose perpendicular distance from O is R, so each gives

$\text B_1 = \dfrac{\text µ_0}{4\text π}\dfrac{\text I}{\text R}(\sin 0° + \sin 90°) \\[1em] \text B_1 = \dfrac{\text µ_0}{4\text π}\dfrac{\text I}{\text R}$

and by the right-hand thumb rule both are directed along $-\hat{\text k}$. Their sum is

$\vec{\text B}_1 = 2\left(\dfrac{\text µ_0}{4\text π}\dfrac{\text I}{\text R}\right)(-\hat{\text k})$

Field due to the semi-circular portion : The field at the centre due to a circular arc of radius R subtending an angle θ is

$\text B = \dfrac{\text µ_0}{4\text π}\dfrac{\text I}{\text R}\text θ$

For a semi-circle, θ = π, so

$\vec{\text B}_2 = \dfrac{\text µ_0 \text I}{4\text R}(-\hat{\text i}) \\[1em] \vec{\text B}_2 = \text π\left(\dfrac{\text µ_0}{4\text π}\dfrac{\text I}{\text R}\right)(-\hat{\text i})$

Resultant field :

$\vec{\text B} = \vec{\text B}_1 + \vec{\text B}_2 \\[1em] \vec{\text B} = -\dfrac{\text µ_0}{4\text π}\dfrac{\text I}{\text R}(\text π\hat{\text i} + 2\hat{\text k})$

Two coaxial solenoids of different radii carry current I in the same direction. Let $\vec{\text F}_1$ be the magnetic force on the inner solenoid due to the outer one and $\vec{\text F}_2$ be the magnetic force on the outer solenoid due to the inner one. Then :

1. $\vec{\text F}_1 = \vec{\text F}_2 = 0$
2. $\vec{\text F}_1$ is radially inwards and $\vec{\text F}_2$ is radially outwards
3. $\vec{\text F}_1$ is radially inwards and $\vec{\text F}_2 = 0$
4. $\vec{\text F}_1$ is radially outwards and $\vec{\text F}_2 = 0$

Answer

$\vec{\text F}_1 = \vec{\text F}_2 = 0$

Reason — The magnetic field of a long solenoid is confined to its interior; outside the solenoid the field is zero.

The outer solenoid therefore lies in a region where the field of the inner solenoid is zero, so $\vec{\text F}_2 = 0$.

The inner solenoid lies inside the outer one, where the field is uniform and directed along the axis. The force on each current-element of a turn is radial, and on going round a complete turn these radial forces cancel in pairs. Hence the total force on the inner solenoid is also zero, $\vec{\text F}_1 = 0$.

Two long current carrying thin wires, both with current I, are held by insulating threads of length L and are in equilibrium as shown in the figure, with threads making an angle θ with the vertical. If wires have mass λ per unit length then the value of I is (g = gravitational acceleration) :

1. $\sin θ\sqrt{\dfrac{\text{πλgL}}{\text µ_0 \cos θ}}$
2. $2\sin θ\sqrt{\dfrac{\text{πλgL}}{\text µ_0 \cos θ}}$
3. $2\sqrt{\dfrac{\text{πgL}}{\text µ_0}\tan θ}$
4. $\sqrt{\dfrac{\text{πλgL}}{\text µ_0}\tan θ}$

Answer

$2\sin θ\sqrt{\dfrac{\text{πλgL}}{\text µ_0 \cos θ}}$

Reason — Given, threads of length L making an angle θ with the vertical, and wires of mass λ per unit length.

The separation between the two wires is x = 2L sin θ, and the currents are in opposite directions, so the force per unit length between them is repulsive,

$\text F = \dfrac{\text µ_0}{2\text π}\dfrac{\text I^2}{2\text L \sin θ} \\[1em] \text F = \dfrac{\text µ_0 \text I^2}{4\text{πL} \sin θ}$

For equilibrium of each wire, resolving the tension T in the thread,

$\text T \sin θ = \text F, \qquad \text T \cos θ = \text{λg}$

Dividing,

$\tan θ = \dfrac{\text F}{\text{λg}} \\[1em] \dfrac{\text µ_0 \text I^2}{4\text{πL} \sin θ} = \text{λg} \tan θ \\[1em] \text I^2 = \dfrac{4\text{πLλg} \sin^2 θ}{\text µ_0 \cos θ} \\[1em] \text I = 2 \sin θ\sqrt{\dfrac{\text{πλgL}}{\text µ_0 \cos θ}}$

A rectangular loop of sides 10 cm and 5 cm carrying a current I of 12 A is placed in different orientations as shown in the figures :

If there is a uniform magnetic field of 0.3 T in the positive Z direction, in which orientations the loop would be in (i) stable equilibrium and (ii) unstable equilibrium?

1. (1) and (2) respectively
2. (1) and (3) respectively
3. (2) and (4) respectively
4. (2) and (3) respectively

Answer

(2) and (4) respectively

Reason — A current loop placed in a magnetic field is in stable equilibrium when its magnetic moment $\vec{\text M} \space$is parallel to $\vec{\text B} \space$, and in unstable equilibrium when $\vec{\text M} \space$is anti-parallel to $\vec{\text B} \space$.

The field is along the positive Z direction. Applying the right-hand palm rule no. 1 to each orientation, the magnetic moment of the loop in orientation (2) points along the positive Z direction, so it is in stable equilibrium, while in orientation (4) it points along the negative Z direction, so it is in unstable equilibrium.

A long straight wire of radius a carries a steady current I. The current is uniformly distributed over its cross-section. The ratio of the magnetic fields B₁ and B₂ at radial distances a/2 and 2a respectively, from the axis of the wire is :

1. 1/4
2. 1/2
3. 1
4. 4

Answer

1

Reason — Given, a wire of radius a carrying a steady current I distributed uniformly over the cross-section.

At the radial distance a/2 (inside the wire) :

$\text B_1 = \dfrac{\text µ_0 \text I}{2\text π}\dfrac{(\text a/2)}{\text a^2} \\[1em] \text B_1 = \dfrac{\text µ_0 \text I}{4\text{πa}}$

At the radial distance 2a (outside the wire) :

$\text B_2 = \dfrac{\text µ_0 \text I}{2\text π(2\text a)} \\[1em] \text B_2 = \dfrac{\text µ_0 \text I}{4\text{πa}} \\[1em] \dfrac{\text B_1}{\text B_2} = 1$

A long straight wire of radius 'a' carries a steady current I. The current is uniformly distributed across its cross-section. The ratio of the magnetic field at a/2 and 2a is :

1. 4
2. 1
3. 1/2
4. 1/4

Answer

1

Reason — Given, a wire of radius a carrying a steady current I distributed uniformly across its cross-section.

Inside the wire, $\text B = \dfrac{\text µ_0 \text{Ir}}{2\text{πa}^2}$, so at r = a/2,

$\text B_1 = \dfrac{\text µ_0 \text I}{2\text{πa}^2} \times \dfrac{\text a}{2} \\[1em] \text B_1 = \dfrac{\text µ_0 \text I}{4\text{πa}}$

Outside the wire, $\text B = \dfrac{\text µ_0 \text I}{2\text{πr}}$, so at r = 2a,

$\text B_2 = \dfrac{\text µ_0 \text I}{4\text{πa}} \\[1em] \dfrac{\text B_1}{\text B_2} = 1$

A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the centre of the loop is B. It is then bent into a circular coil of n turns. The magnetic field at the centre of this coil of n turns will be :

1. 2n²B
2. nB
3. n²B
4. 2nB

Answer

n²B

Reason — Let the length of the wire be L.

For one turn : L = 2πr₁, so $\text r_1 = \dfrac{\text L}{2\text π}$ and

$\text B = \dfrac{\text µ_0 \text I}{2\text r_1} \\[1em] \text B = \dfrac{\text µ_0 \text{πI}}{\text L}$

For n turns : L = n(2πr₂), so $\text r_2 = \dfrac{\text L}{2\text{πn}}$ and

$\text B' = \dfrac{\text µ_0 \text n \text I}{2\text r_2} \\[1em] \text B' = \dfrac{\text µ_0 \text n \text I \times 2\text{πn}}{2\text L} \\[1em] \text B' = \dfrac{\text µ_0 \text{πn}^2 \text I}{\text L} \\[1em] \text B' = \text n^2 \text B$

Two identical wires A and B, each of length 'l' carry the same current I. Wire A is bent into a circle of radius R and wire B is bent to form a square of side 'a'. If B_A and B_B are the values of magnetic field at the centres of the circle and square respectively, then the ratio B_A/B_B is :

1. $\dfrac{\text π^2}{16\sqrt{2}}$
2. $\dfrac{\text π^2}{16}$
3. $\dfrac{\text π^2}{8\sqrt{2}}$
4. $\dfrac{\text π^2}{8}$

Answer

$\dfrac{\text π^2}{8\sqrt{2}}$

Reason — Given, each wire has length l and carries the same current I.

Wire A, bent into a circle : l = 2πR, so $\text R = \dfrac{l}{2\text π}$ and

$\text B_\text A = \dfrac{\text µ_0 \text I}{2\text R} \\[1em] \text B_\text A = \dfrac{\text µ_0 \text{πI}}{l}$

Wire B, bent into a square : the side is $\text a = \dfrac{l}{4}$, and the perpendicular distance of each side from the centre is $\dfrac{\text a}{2} = \dfrac{l}{8}$. For one side, φ₁ = φ₂ = 45°, so

$\text B_{\text{side}} = \dfrac{\text µ_0}{4\text π}\dfrac{\text I}{(l/8)}(\sin 45° + \sin 45°) \\[1em] \text B_{\text{side}} = \dfrac{2\text µ_0 \text I}{\text π l} \times \sqrt{2}$

All the four sides give fields in the same direction, so

$\text B_\text B = 4 \times \dfrac{2\sqrt{2}\text µ_0 \text I}{\text π l} \\[1em] \text B_\text B = \dfrac{8\sqrt{2}\text µ_0 \text I}{\text π l} \\[1em] \dfrac{\text B_\text A}{\text B_\text B} = \dfrac{\text µ_0 \text{πI}/l}{8\sqrt{2}\text µ_0 \text I/\text π l} \\[1em] \dfrac{\text B_\text A}{\text B_\text B} = \dfrac{\text π^2}{8\sqrt{2}}$

An electron is moving in a circular path under the influence of a transverse magnetic field 3.57 × 10^-2 T. If the value of e/m is 1.76 × 10¹¹ C/kg, the frequency of revolution of the electron is :

1. 6.82 MHz
2. 1 GHz
3. 100 MHz
4. 62.8 MHz

Answer

1 GHz

Reason — Given, B = 3.57 × 10^-2 T and $\dfrac{\text e}{\text m}$ = 1.76 × 10¹¹ C/kg.

The frequency of revolution of a charged particle in a magnetic field is

$\text ν = \dfrac{\text{eB}}{2\text{πm}} \\[1em] \text ν = \dfrac{1}{2\text π}\left(\dfrac{\text e}{\text m}\right)\text B$

Substituting the values,

$\text ν = \dfrac{(1.76 \times 10^{11})(3.57 \times 10^{-2})}{2 \times 3.14} \\[1em] \text ν = \dfrac{6.283 \times 10^9}{6.28} \\[1em] \text ν = 1 \times 10^9\ \text{Hz} = 1\ \text{GHz}$

A square loop ABCD carrying a current i (see figure), is placed near and coplanar with a long straight conductor XY carrying a current I, the net force on the loop will be :

1. $\dfrac{2\text µ_0 \text{Ii}}{3\text π}$
2. $\dfrac{\text µ_0 \text{Ii}}{2\text π}$
3. $\dfrac{2\text µ_0 \text{IiL}}{3\text π}$
4. $\dfrac{\text µ_0 \text{IiL}}{2\text π}$

Answer

$\dfrac{2\text µ_0 \text{Ii}}{3\text π}$

Reason — Given, a square loop ABCD of side L carrying a current i, placed coplanar with a long straight conductor XY carrying a current I, the nearer side being at a distance $\dfrac{\text L}{2}$ from XY.

The nearer side AB is at a distance $\dfrac{\text L}{2}$ and the farther side DC at a distance $\dfrac{\text L}{2} + \text L = \dfrac{3\text L}{2}$.

Force on the nearer side :

$\text F_1 = \dfrac{\text µ_0}{2\text π}\dfrac{\text{Ii}}{(\text L/2)} \times \text L \\[1em] \text F_1 = \dfrac{\text µ_0 \text{Ii}}{\text π}$

Force on the farther side :

$\text F_2 = \dfrac{\text µ_0}{2\text π}\dfrac{\text{Ii}}{(3\text L/2)} \times \text L \\[1em] \text F_2 = \dfrac{\text µ_0 \text{Ii}}{3\text π}$

The currents in these two sides are oppositely directed, so one force is an attraction and the other a repulsion. The forces on the other two sides are equal and opposite and cancel. Hence the net force is

$\text F = \text F_1 - \text F_2 \\[1em] \text F = \dfrac{\text µ_0 \text{Ii}}{\text π} - \dfrac{\text µ_0 \text{Ii}}{3\text π} \\[1em] \text F = \dfrac{2\text µ_0 \text{Ii}}{3\text π}$

A symmetric star shaped conducting wire loop is carrying a steady state current I as shown in the figure. The distance between the diametrically opposite vertices of the star is 4a. The magnitude of the magnetic field at the centre of the loop is :

1. $\dfrac{\text µ_0 \text I}{4\text{πa}}3[2 - \sqrt{3}]$
2. $\dfrac{\text µ_0 \text I}{4\text{πa}}6[\sqrt{3} - 1]$
3. $\dfrac{\text µ_0 \text I}{4\text{πa}}3[\sqrt{3} - 1]$
4. $\dfrac{\text µ_0 \text I}{4\text{πa}}6[\sqrt{3} + 1]$

Answer

$\dfrac{\text µ_0 \text I}{4\text{πa}}6[\sqrt{3} - 1]$

Reason — Given, a symmetric star shaped loop carrying a current I, the distance between the diametrically opposite vertices being 4a.

The star is made up of 12 straight segments, and by symmetry each of them produces the same magnitude of field at the centre, all in the same direction.

For each segment the perpendicular distance from the centre is d = OA = a, and its ends subtend 60° and 30° at O on the same side of the normal OA (that is, φ₁ = 60°, φ₂ = −30°), so

$\text B_{\text{seg}} = \dfrac{\text µ_0}{4\text π}\dfrac{\text I}{\text d}(\sin \text φ_1 + \sin \text φ_2)$

Adding the contributions of all the twelve segments,

$\text B = 12 \times \dfrac{\text µ_0}{4\text π}\dfrac{\text I}{\text a}[\sin 60° + \sin(-30°)] \\[1em] \text B = 12 \times \dfrac{\text µ_0 \text I}{4\text{πa}}\left(\dfrac{\sqrt{3}}{2} - \dfrac{1}{2}\right) \\[1em] \text B = \dfrac{\text µ_0 \text I}{4\text{πa}}6[\sqrt{3} - 1]$

An arrangement of three parallel straight wires placed perpendicular to plane of paper carrying same current I along the same direction is shown in Fig. Magnitude of force per unit length on the middle wire 'B' is given by :

1. $\dfrac{\text µ_0 \text I^2}{2\text{πd}}$
2. $\dfrac{2\text µ_0 \text I^2}{\text{πd}}$
3. $\dfrac{\sqrt{2}\text µ_0 \text I^2}{\text{πd}}$
4. $\dfrac{\text µ_0 \text I^2}{\sqrt{2}\text{πd}}$

Answer

$\dfrac{\text µ_0 \text I^2}{\sqrt{2}\text{πd}}$

Reason — Given, three parallel wires carrying the same current I along the same direction, the wire B being at a distance d from A and d from C, with AB perpendicular to BC.

The force per unit length on B due to A is

$\text F_1 = \dfrac{\text µ_0}{2\text π}\dfrac{\text I^2}{\text d}$

and that due to C is

$\text F_2 = \dfrac{\text µ_0}{2\text π}\dfrac{\text I^2}{\text d}$

Since all the currents are in the same direction, B is attracted towards A and towards C. The angle between BA and BC is 90°, so the two forces are at right angles and the resultant is

$\text F = \sqrt{\text F_1^2 + \text F_2^2} \\[1em] \text F = \sqrt{2}\left(\dfrac{\text µ_0 \text I^2}{2\text{πd}}\right) \\[1em] \text F = \dfrac{\text µ_0 \text I^2}{\sqrt{2}\ \text{πd}}$

A metallic rod of mass per unit length 0.5 kg m^-1 is lying horizontally on a smooth inclined plane which makes an angle of 30° with the horizontal. The rod is not allowed to slide down by flowing current through it when a magnetic field of induction 0.25 T is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is :

1. 7.14 A
2. 11.32 A
3. 14.76 A
4. 5.98 A

Answer

11.32 A

Reason — Given,

• mass per unit length, $\dfrac{\text m}{l}$ = 0.5 kg m^-1
• θ = 30°, B = 0.25 T, g = 9.8 m s^-2

The rod lies horizontally on the smooth inclined plane and the field is vertical, so the magnetic force IBl is horizontal. Balancing the components along the inclined plane,

$\text{IB}l \cos θ = \text{mg} \sin θ \\[1em] \text I = \dfrac{\text m}{l} \times \dfrac{\text g \tan θ}{\text B} \\[1em] \text I = 0.5 \times \dfrac{(9.8)(\tan 30°)}{0.25} \\[1em] \text I = 0.5 \times \dfrac{(9.8)(0.5774)}{0.25} \\[1em] \text I = \dfrac{2.829}{0.25} \\[1em] \text I = 11.32\ \text A$

An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbits of radii r_e, r_p, r_α respectively in uniform magnetic field B. The relation between r_e, r_p, r_α is :

1. r_e < r_p < r_α
2. r_e < r_α < r_p
3. r_e > r_p = r_α
4. r_e < r_p = r_α

Answer

r_e < r_p = r_α

Reason — Given, the three particles have the same kinetic energy K in the same field B.

$\text r = \dfrac{\sqrt{2\text{mK}}}{\text{qB}} \quad \Rightarrow \quad \text r \propto \dfrac{\sqrt{\text m}}{\text q}$

Taking the proton mass as m and its charge as e,

• electron : $\text r_\text e \propto \dfrac{\sqrt{\text m_\text e}}{\text e}$, and m_e is very much smaller than m
• proton : $\text r_\text p \propto \dfrac{\sqrt{\text m}}{\text e}$
• α-particle : $\text r_α \propto \dfrac{\sqrt{4\text m}}{2\text e} = \dfrac{\sqrt{\text m}}{\text e}$

Hence r_p = r_α, and since m_e < m, r_e < r_p = r_α.

As shown in the figure, two infinitely long, identical wires are bent by 90° and placed in such a way that the segments LP and QM are along the x-axis, while segments PS and QN are parallel to the y-axis. If OP = OQ = 4 cm, and the magnitude of the magnetic field at O is 10^-4 T, and the two wires carry equal currents (see figure), the magnitude of the current in each wire and the direction of the magnetic field at O will be (µ₀ = 4π × 10^-7 NA^-2) :

1. 40 A, perpendicular into the page
2. 40 A, perpendicular out of the page
3. 20 A, perpendicular out of the page
4. 20 A, perpendicular into the page

Answer

20 A, perpendicular into the page

Reason — Given, OP = OQ = 4 cm = 0.04 m and B = 10^-4 T at O.

The segments LP and QM lie along the x-axis, and O also lies on this axis. For these segments θ = 0 or 180°, so they develop no magnetic field at O.

The segments PS and QN are semi-infinite conductors, each at a perpendicular distance 0.04 m from O. Each therefore gives

$\text B_1 = \dfrac{\text µ_0}{4\text π}\dfrac{\text I}{\text d}(\sin 0° + \sin 90°) \\[1em] \text B_1 = \dfrac{\text µ_0}{4\text π}\dfrac{\text I}{\text d}$

Applying the right-hand thumb rule, both these fields at O are directed into the page, so

$\text B = 2 \times \dfrac{\text µ_0}{4\text π}\dfrac{\text I}{\text d} \\[1em] \text B = \dfrac{\text µ_0 \text I}{2\text{πd}} \\[1em] 10^{-4} = (2 \times 10^{-7}) \times \dfrac{\text I}{0.04} \\[1em] \text I = \dfrac{(10^{-4})(0.04)}{2 \times 10^{-7}} \\[1em] \text I = 20\ \text A$

Two very long, straight, and insulated wires are kept at 90° angle from each other in xy-plane as shown in the figure. These wires carry currents of equal magnitude I, whose directions are shown in the figure. The net magnetic field at point P will be :

1. zero
2. $\dfrac{+\text µ_0 \text I}{\text{πd}}(\hat{\text z})$
3. $-\dfrac{\text µ_0 \text I}{2\text{πd}}(\hat{\text x} + \hat{\text y})$
4. $\dfrac{\text µ_0 \text I}{2\text{πd}}(\hat{\text x} + \hat{\text y})$

Answer

zero

Reason — Given, two mutually perpendicular wires carrying equal currents I, and the point P at a perpendicular distance d from each of them.

The field at P due to each wire has the magnitude

$\text B_1 = \text B_2 = \dfrac{\text µ_0 \text I}{2\text{πd}}$

Applying the right-hand thumb rule to the two wires for the directions of current shown, the field due to one wire at P is directed into the plane of the page while that due to the other is directed out of the plane of the page. Being equal in magnitude and opposite in direction, they cancel,

$\vec{\text B} = \vec{\text B}_1 + \vec{\text B}_2 \\[1em] \vec{\text B} = 0$

A cylindrical conductor of radius R is carrying a constant current. The plot of the magnitude of the magnetic field. B with the distance d from the centre of the conductor, is correctly represented by the figure :

Answer

Reason — For a cylindrical conductor of radius R carrying a constant current distributed uniformly over its cross-section,

inside the conductor (d < R) :

$\text B = \dfrac{\text µ_0 \text I}{2\text π}\dfrac{\text d}{\text R^2} \quad \Rightarrow \quad \text B \propto \text d$

so B increases linearly with d, reaching its maximum value $\dfrac{\text µ_0 \text I}{2\text{πR}}$ at the surface;

outside the conductor (d > R) :

$\text B = \dfrac{\text µ_0 \text I}{2\text{πd}} \quad \Rightarrow \quad \text B \propto \dfrac{1}{\text d}$

so B decreases hyperbolically. This variation is represented by the first graph.

A thin ring of 10 cm radius carries a uniformly distributed charge. The ring rotates at a constant angular speed of 40π rads^-1 about its axis, perpendicular to its plane. If the magnetic field at its centre is 3.8 × 10^-9 T, then the charge carried by the ring is close to (µ₀ = 4π × 10^-7 N/A²).

1. 7 × 10^-6 C
2. 4 × 10^-5 C
3. 2 × 10^-6 C
4. 3 × 10^-5 C

Answer

3 × 10^-5 C

Reason — Given,

• r = 10 cm = 0.1 m
• ω = 40π rad s^-1
• B = 3.8 × 10^-9 T

The rotating ring is equivalent to a current loop. If q is the charge on the ring, the equivalent current is

$\text I = \dfrac{\text q}{\text T} \\[1em] \text I = \dfrac{\text{qω}}{2\text π} \\[1em] \text I = \dfrac{\text q(40\text π)}{2\text π} \\[1em] \text I = 20\text q$

The field at the centre of the ring is

$\text B = \dfrac{\text µ_0 \text I}{2\text r} \\[1em] \text B = \dfrac{(4\text π \times 10^{-7})(20\text q)}{2 \times 0.1} \\[1em] 3.8 \times 10^{-9} = (1.2566 \times 10^{-4})\text q \\[1em] \text q = \dfrac{3.8 \times 10^{-9}}{1.2566 \times 10^{-4}} \\[1em] \text q = 3 \times 10^{-5}\ \text C$

One of the two identical conducting wires of length L is bent in the form of a circular loop and the other one into a circular coil of N identical turns. If the same current is passed in both, the ratio of the magnetic field at the central of the loop (B_L) to that at the centre of the coil (B_C), i.e., $\text R\dfrac{\text B_\text L}{\text B_\text C}$ Will be :

1. $\dfrac{1}{\text N}$
2. N²
3. $\dfrac{1}{\text N^2}$
4. N

Answer

$\dfrac{1}{\text N^2}$

Reason — Given, both the wires have the same length L and carry the same current.

For the single loop : L = 2πR_L, so $\text R_\text L = \dfrac{\text L}{2\text π}$ and

$\text B_\text L = \dfrac{\text µ_0 \text I}{2\text R_\text L} \\[1em] \text B_\text L = \dfrac{\text µ_0 \text{πI}}{\text L}$

For the coil of N turns : L = N(2πR_C), so $\text R_\text C = \dfrac{\text L}{2\text{πN}}$ and

$\text B_\text C = \dfrac{\text µ_0 \text N \text I}{2\text R_\text C} \\[1em] \text B_\text C = \dfrac{\text µ_0 \text{πN}^2 \text I}{\text L} \\[1em] \dfrac{\text B_\text L}{\text B_\text C} = \dfrac{1}{\text N^2}$

Find the magnetic field at point P due to a straight line segment AB of length 6 cm carrying a current of 5A. (see figure) (µ₀ = 4π × 10^-7 N-A^-2)

1. 1.5 × 10^-5 T
2. 3.0 × 10^-5 T
3. 2.0 × 10^-5 T
4. 2.5 × 10^-5 T

Answer

1.5 × 10^-5 T

Reason — Given, AB = 6 cm, PA = PB = 5 cm and I = 5 A.

The perpendicular distance of P from AB is

$\text d = \sqrt{(5)^2 - (3)^2} \\[1em] \text d = \sqrt{25 - 9} \\[1em] \text d = 4\ \text{cm} = 0.04\ \text m$

For a straight conductor of finite length,

$\text B = \dfrac{\text µ_0}{4\text π}\dfrac{\text I}{\text d}(\sin \text φ_1 + \sin \text φ_2)$

Here, by symmetry, $\sin \text φ_1 = \sin \text φ_2 = \dfrac{3}{5} = 0.6$. Substituting,

$\text B = (10^{-7}) \times \dfrac{5}{0.04} \times (0.6 + 0.6) \\[1em] \text B = (10^{-7})(125)(1.2) \\[1em] \text B = 1.5 \times 10^{-5}\ \text T$

A particle of mass m and charge q is in an electric and magnetic field given by $\vec{\text E} = 2\hat{\text i} + 3\hat{\text j}$; $\vec{\text B} = 4\hat{\text j} + 6\hat{\text k}$. The charged particle is shifted from the origin to the point P(x = 1; y = 1) along a straight path. The magnitude of the total work done is :

1. (0.35)q
2. (0.15)q
3. (2.5)q
4. 5q

Answer

5q

Reason — Given, $\vec{\text E} = 2\hat{\text i} + 3\hat{\text j}$ and the displacement from the origin to P(1, 1) is $\vec{\text d} = \hat{\text i} + \hat{\text j}$.

The magnetic force is always perpendicular to the velocity, so the work done by the magnetic field is zero. Only the electric force does work,

$\text W = \text q\vec{\text E} \cdot \vec{\text d} \\[1em] \text W = \text q(2\hat{\text i} + 3\hat{\text j}) \cdot (\hat{\text i} + \hat{\text j}) \\[1em] \text W = \text q(2 + 3) \\[1em] \text W = 5\text q$

A rigid square loop of side 'a' and carrying current I₂ is lying on a horizontal surface near a long current I₁ carrying wire in the same plane as shown in figure. The net force on the loop due to wire will be :

1. attractive and equal to $\dfrac{\text µ_0 \text I_1 \text I_2}{3\text π}$
2. repulsive and equal to $\dfrac{\text µ_0 \text I_1 \text I_2}{4\text π}$
3. repulsive and equal to $\dfrac{\text µ_0 \text I_1 \text I_2}{2\text π}$
4. zero

Answer

repulsive and equal to $\dfrac{\text µ_0 \text I_1 \text I_2}{4\text π}$

Reason — Given, a square loop of side a carrying a current I₂, the nearer side being at a distance a from the long wire carrying I₁.

Force on the nearer side (at a distance a) :

$\text F_1 = \dfrac{\text µ_0}{2\text π}\dfrac{\text I_1 \text I_2}{\text a} \times \text a \\[1em] \text F_1 = \dfrac{\text µ_0 \text I_1 \text I_2}{2\text π}$

Force on the farther side (at a distance 2a) :

$\text F_2 = \dfrac{\text µ_0}{2\text π}\dfrac{\text I_1 \text I_2}{2\text a} \times \text a \\[1em] \text F_2 = \dfrac{\text µ_0 \text I_1 \text I_2}{4\text π}$

The forces on the other two sides are equal and opposite and cancel out. The current in the nearer side is opposite to I₁, so F₁ is a repulsion, while F₂ is an attraction. Hence the net force is

$\text F = \text F_1 - \text F_2 \\[1em] \text F = \dfrac{\text µ_0 \text I_1 \text I_2}{2\text π} - \dfrac{\text µ_0 \text I_1 \text I_2}{4\text π} \\[1em] \text F = \dfrac{\text µ_0 \text I_1 \text I_2}{4\text π}$

and it is repulsive, since the larger force is the repulsive one.

An electron, moving along the x-axis with an initial energy of 100 eV, enters a region of magnetic field $\vec{\text B} = (1.5 \times 10^{-3} \text{ T})\hat{\text k}$ at S (see figure). The field extends between x = 0 and x = 2 cm. The electron is detected at the point Q on a screen placed 8 cm away from the point S. The distance d between P and Q (on the screen) is :

(Electron's charge = 1.6 × 10^-19 C, mass of electron = 9.1 × 10^-31 kg)

1. 2.25 cm
2. 12.87 cm
3. 1.22 cm
4. 11.65 cm

Answer

12.87 cm

Reason — Given, energy = 100 eV, B = 1.5 × 10^-3 T, the field extending over x = 2 cm, and the screen 8 cm from S.

Since $\vec{\text B} \space$is along $+\hat{\text k}$, the field points out of the page (the crosses in the question figure do not match the stated $\hat{\text k}$), and the force on the electron, $-\text e(\vec{\text v} \times \vec{\text B})$, is along +y, so it bends towards Q.

Speed of the electron :

$\text v = \sqrt{\dfrac{2\text K}{\text m}} \\[1em] \text v = \sqrt{\dfrac{2(100)(1.6 \times 10^{-19})}{9.1 \times 10^{-31}}} \\[1em] \text v = 5.93 \times 10^6\ \text{ms}^{-1}$

Radius of the circular path inside the field :

$\text r = \dfrac{\text{mv}}{\text{eB}} \\[1em] \text r = \dfrac{(9.1 \times 10^{-31})(5.93 \times 10^6)}{(1.6 \times 10^{-19})(1.5 \times 10^{-3})} \\[1em] \text r = 2.25\ \text{cm}$

Deflection inside the field region : with x = 2 cm,

$\text y_1 = \text r - \sqrt{\text r^2 - \text x^2} \\[1em] \text y_1 = 2.25 - \sqrt{(2.25)^2 - (2)^2} \\[1em] \text y_1 = 2.25 - 1.03 \\[1em] \text y_1 = 1.22\ \text{cm}$

Deflection outside the field region : the electron leaves the field at an angle θ with

$\sin θ = \dfrac{\text x}{\text r} \\[1em] \sin θ = \dfrac{2}{2.25} \\[1em] \sin θ = 0.889 \\[1em] \cos θ = 0.458 \\[1em] \tan θ = 1.94$

and then travels in a straight line over the remaining 8 − 2 = 6 cm,

$\text y_2 = 6 \tan θ \\[1em] \text y_2 = 6 \times 1.94 \\[1em] \text y_2 = 11.64\ \text{cm}$

Total distance :

$\text d = \text y_1 + \text y_2 \\[1em] \text d = 1.22 + 11.64 \\[1em] \text d = 12.86\ \text{cm}$

This agrees with option 2 (12.87 cm) to within rounding; with the unrounded radius (2.248 cm), d ≈ 12.9 cm.

Two wires A and B are carrying currents I₁ and I₂ as shown in the figure. The separation between them is d. A third wire C carrying a current I is to be kept parallel to them at a distance x from A such that the net force acting on it is zero. The possible values of x are :

1. $\text x = \left(\dfrac{\text I_1}{\text I_1 - \text I_2}\right)\text d$ and $\text x = \dfrac{\text I_2}{(\text I_1 + \text I_2)}\text d$
2. $\text x = \pm\dfrac{\text I_1 \text d}{(\text I_1 - \text I_2)}$
3. $\text x = \left(\dfrac{\text I_1}{\text I_1 + \text I_2}\right)\text d$ and $\text x = \dfrac{\text I_2}{(\text I_1 - \text I_2)}\text d$
4. $\text x = \left(\dfrac{\text I_2}{\text I_1 + \text I_2}\right)\text d$ and $\text x = \left(\dfrac{\text I_2}{\text I_1 - \text I_2}\right)\text d$

Answer

$\text x = \pm\dfrac{\text I_1 \text d}{(\text I_1 - \text I_2)}$

Reason — Given, the wires A and B carry currents I₁ and I₂ in opposite directions, separated by a distance d, and the wire C is at a distance x from A.

The net force on C is zero when the two forces per unit length are equal and opposite. Since the currents in A and B are oppositely directed, the two forces act in the same direction at every point between the wires, so no null point can lie between them; C must lie outside.

C beyond B (distance x − d from B) :

$\dfrac{\text µ_0}{2\text π}\dfrac{\text I_1 \text I}{\text x} = \dfrac{\text µ_0}{2\text π}\dfrac{\text I_2 \text I}{(\text x - \text d)}$

Solving,

$\text I_1(\text x - \text d) = \text I_2 \text x \\[1em] \text x(\text I_1 - \text I_2) = \text I_1 \text d \\[1em] \text x = \dfrac{\text I_1 \text d}{\text I_1 - \text I_2}$

C beyond A (distance x + d from B) :

$\dfrac{\text I_1}{\text x} = \dfrac{\text I_2}{\text x + \text d} \\[1em] \text x = \dfrac{\text I_1 \text d}{\text I_2 - \text I_1} \\[1em] \text x = -\dfrac{\text I_1 \text d}{\text I_1 - \text I_2}$

Combining the two cases,

$\text x = \pm\dfrac{\text I_1 \text d}{(\text I_1 - \text I_2)}$

Ionized hydrogen atoms and α-particles with same momenta enters perpendicular to a constant magnetic field B. The ratio of their radii of their paths r_H : r_α will be :

1. 4 : 1
2. 1 : 4
3. 2 : 1
4. 1 : 2

Answer

2 : 1

Reason — Given, the ionized hydrogen atoms and the α-particles have the same momentum p and enter the same field B perpendicularly.

$\text r = \dfrac{\text p}{\text{qB}} \quad \Rightarrow \quad \text r \propto \dfrac{1}{\text q}$

The charge on an ionized hydrogen atom is e and that on an α-particle is 2e, so

$\dfrac{\text r_\text H}{\text r_α} = \dfrac{\text q_α}{\text q_\text H} \\[1em] \dfrac{\text r_\text H}{\text r_α} = \dfrac{2\text e}{\text e} \\[1em] \dfrac{\text r_\text H}{\text r_α} = \dfrac{2}{1}$

A proton, an electron and a helium nucleus, have the same energy. They are in circular orbits in a plane due to magnetic field perpendicular to the plane. Let r_p, r_e and r_He be their respective radii, then :

1. r_e < r_p < r_He
2. r_e > r_p = r_He
3. r_e < r_p = r_He
4. r_e > r_p > r_He

Answer

r_e < r_p = r_He

Reason — Given, the three particles have the same energy K in the same magnetic field B.

$\text r = \dfrac{\sqrt{2\text{mK}}}{\text{qB}} \quad \Rightarrow \quad \text r \propto \dfrac{\sqrt{\text m}}{\text q}$

Taking the proton mass as m and its charge as e,

• proton : $\text r_\text p \propto \dfrac{\sqrt{\text m}}{\text e}$
• helium nucleus : m_He = 4m and q_He = 2e, so $\text r_{\text{He}} \propto \dfrac{\sqrt{4\text m}}{2\text e} = \dfrac{\sqrt{\text m}}{\text e}$
• electron : $\text r_\text e \propto \dfrac{\sqrt{\text m_\text e}}{\text e}$, and m_e is very much smaller than m

Hence r_p = r_He, while r_e is much smaller, giving r_e < r_p = r_He.

A particle having the same charge as of electron moves in a circular path of radius 0.5 cm under the influence of a magnetic field of 0.5 T. If an electric field of 100 V/m makes it to move in a straight path, then the mass of the particle is : (Given charge of electron = 1.6 × 10^-19 C)

1. 2.0 × 10^-24 kg
2. 1.6 × 10^-19 kg
3. 1.6 × 10^-27 kg
4. 9.1 × 10^-31 kg

Answer

2.0 × 10^-24 kg

Reason — Given,

• r = 0.5 cm = 5 × 10^-3 m
• B = 0.5 T
• E = 100 V/m
• q = 1.6 × 10^-19 C

The particle moves in a straight path when the electric force balances the magnetic force,

$\text{qE} = \text{qvB} \\[1em] \text v = \dfrac{\text E}{\text B} \\[1em] \text v = \dfrac{100}{0.5} \\[1em] \text v = 200\ \text{ms}^{-1}$

From the radius of the circular path, $\text r = \dfrac{\text{mv}}{\text{qB}}$,

$\text m = \dfrac{\text{qBr}}{\text v} \\[1em] \text m = \dfrac{(1.6 \times 10^{-19})(0.5)(5 \times 10^{-3})}{200} \\[1em] \text m = \dfrac{4 \times 10^{-22}}{200} \\[1em] \text m = 2.0 \times 10^{-24}\ \text{kg}$

A particle of charge q and mass m is moving with a velocity $-\text v\hat{\text i}$ (v ≠ 0) towards a large screen placed in the Y-Z plane at a distance d. If there is a magnetic field $\vec{\text B} = \text B_0\hat{\text k}$, the maximum value of v for which the particle will not hit the screen is :

1. $\dfrac{2\text{qdB}_0}{\text m}$
2. $\dfrac{\text{qdB}_0}{3\text m}$
3. $\dfrac{\text{qdB}_0}{2\text m}$
4. $\dfrac{\text{qdB}_0}{\text m}$

Answer

$\dfrac{\text{qdB}_0}{\text m}$

Reason — Given, $\vec{\text v} = -\text v\hat{\text i}$, $\vec{\text B} = \text B_0\hat{\text k}$ and the screen at a distance d in the Y-Z plane.

The velocity is perpendicular to the field, so the particle describes a circle of radius

$\text r = \dfrac{\text{mv}}{\text{qB}_0}$

Starting along the −X direction, the greatest distance the particle can penetrate along its initial direction of motion is equal to the radius r of the circle, after which it turns back.

The particle will not hit the screen so long as

$\text r \lt \text d \quad \Rightarrow \quad \dfrac{\text{mv}}{\text{qB}_0} \lt \text d \\[1em] \text v \lt \dfrac{\text{qdB}_0}{\text m}$

Hence the maximum value of v is $\dfrac{\text{qdB}_0}{\text m}$.

The figure shows a region of length 'l' with a uniform magnetic field of 0.3 T in it and a proton entering the region with velocity 4 × 10⁵ ms^-1 making an angle 60° with the field. If the proton completes 10 revolutions by the time it cross the region shown, 'l' is close to (mass of proton = 1.67 × 10^-27 kg, charge of the proton = 1.6 × 10^-19 C).

1. 0.22 m
2. 0.11 m
3. 0.88 m
4. 0.44 m

Answer

0.44 m

Reason — Given, B = 0.3 T, v = 4 × 10⁵ ms^-1, θ = 60°, m = 1.67 × 10^-27 kg and q = 1.6 × 10^-19 C.

The pitch of the helical path is the distance advanced along the field in one complete revolution,

$\text P = \text v \cos θ \times \dfrac{2\text{πm}}{\text{qB}} \\[1em] \text P = (4 \times 10^5)(\cos 60°) \times \dfrac{2 \times 3.14 \times 1.67 \times 10^{-27}}{(1.6 \times 10^{-19})(0.3)} \\[1em] \text P = (2 \times 10^5) \times \dfrac{1.049 \times 10^{-26}}{4.8 \times 10^{-20}} \\[1em] \text P = 0.0437\ \text m$

The proton completes 10 revolutions while crossing the region, so

$l = 10 \times \text P \\[1em] l = 10 \times 0.0437 \\[1em] l = 0.44\ \text m$

A beam of protons with speed 4 × 10⁵ ms^-1 enters a uniform magnetic field of 0.3 T at an angle of 60° to the magnetic field. The pitch of the resulting helical path of protons is close to (Mass of the proton = 1.67 × 10^-27 kg, charge of the proton = 1.69 × 10^-19 C).

1. 2 cm
2. 12 cm
3. 5 cm
4. 4 cm

Answer

4 cm

Reason — Given, v = 4 × 10⁵ ms^-1, B = 0.3 T, θ = 60°, m = 1.67 × 10^-27 kg and q = 1.69 × 10^-19 C.

The pitch of the helical path is

$\text P = \text v \cos θ \times \dfrac{2\text{πm}}{\text{qB}}$

Substituting the values,

$\text P = (4 \times 10^5)(0.5) \times \dfrac{2 \times 3.14 \times 1.67 \times 10^{-27}}{(1.69 \times 10^{-19})(0.3)} \\[1em] \text P = (2 \times 10^5) \times \dfrac{1.049 \times 10^{-26}}{5.07 \times 10^{-20}} \\[1em] \text P = 0.0414\ \text m \\[1em] \text P \approx 4\ \text{cm}$

An electron is moving in +X direction with a velocity of 6 × 10⁶ ms^-1. It enters a region of uniform electric field of 300 V/cm pointing along +Y direction. The magnitude and direction of the magnetic field set up in this region such that the electron keeps moving along the X direction will be :

1. 5 × 10^-3 T, along +Z direction
2. 5 × 10^-3 T, along −Z direction
3. 3 × 10^-4 T, along +Z direction
4. 3 × 10^-4 T, along −Z direction

Answer

5 × 10^-3 T, along +Z direction

Reason — Given, v = 6 × 10⁶ ms^-1 along +X, and E = 300 V/cm = 3 × 10⁴ V/m along +Y.

The electron keeps moving along X, so the electric and magnetic forces on it balance,

$\text{eE} = \text{evB} \\[1em] \text B = \dfrac{\text E}{\text v} \\[1em] \text B = \dfrac{3 \times 10^4}{6 \times 10^6} \\[1em] \text B = 5 \times 10^{-3}\ \text T$

Direction : The electron carries a charge −e, so the electric force on it is along −Y. The magnetic force must therefore be along +Y. Taking $\vec{\text B} = \text B\hat{\text k}$,

$\vec{\text F} = (-\text e)(\text v\hat{\text i} \times \text B\hat{\text k}) \\[1em] \vec{\text F} = (-\text e)(-\text{vB}\hat{\text j}) \\[1em] \vec{\text F} = \text{evB}\hat{\text j}$

which is along +Y, as required. Hence the field is directed along the +Z direction.

An infinitely long straight conductor carries a current of 5 A as shown. An electron is moving with a speed of 10⁵ m/s parallel to the conductor. The perpendicular distance between the electron and the conductor is 20 cm at an instant. Calculate the magnitude of the force experienced by the electron at that instant.

1. 8 × 10^-20 N
2. 4 × 10^-20 N
3. 8π × 10^-20 N
4. 4π × 10^-20 N

Answer

8 × 10^-20 N

Reason — Given, I = 5 A, v = 10⁵ m/s and r = 20 cm = 0.2 m.

The field produced by the conductor at the position of the electron is

$\text B = \dfrac{\text µ_0}{4\text π}\dfrac{2\text I}{\text r} \\[1em] \text B = (10^{-7}) \times \dfrac{2 \times 5}{0.2} \\[1em] \text B = 5 \times 10^{-6}\ \text T$

The electron moves parallel to the conductor, so it moves at right angles to this field and

$\text F = \text{evB} \\[1em] \text F = (1.6 \times 10^{-19})(10^5)(5 \times 10^{-6}) \\[1em] \text F = 8 \times 10^{-20}\ \text N$

A thick current-carrying cable of radius 'R' carries current 'I' uniformly distributed across its cross-section. The variation of magnetic field B(r) due to the cable with the distance 'r' from the axis of the cable is represented by :

Answer

Reason — For a thick cable of radius R carrying a current I uniformly distributed over its cross-section,

inside the cable (r < R) : the current enclosed by a circle of radius r is $\text I\dfrac{\text r^2}{\text R^2}$, so

$\text B = \dfrac{\text µ_0 \text{Ir}}{2\text{πR}^2} \quad \Rightarrow \quad \text B \propto \text r$

outside the cable (r > R) : the whole current is enclosed, so

$\text B = \dfrac{\text µ_0 \text I}{2\text{πr}} \quad \Rightarrow \quad \text B \propto \dfrac{1}{\text r}$

Hence B increases linearly with r up to the surface of the cable, where it reaches its maximum value, and then falls off hyperbolically. This variation is represented by the fourth graph.

In the product

$\vec{\text F} = \text q(\vec{\text v} \times \vec{\text B}) = \text q\vec{\text v} \times (\text B\hat{\text i} + \text B\hat{\text j} + \text B_0\hat{\text k})$

For q = 1 and $\vec{\text v} = 2\hat{\text i} + 4\hat{\text j} + 6\hat{\text k}$

and $\vec{\text F} = 4\hat{\text i} - 20\hat{\text j} + 12\hat{\text k}$

What will be the complete expression for $\vec{\text B} \space$?

1. $6\hat{\text i} + 6\hat{\text j} - 8\hat{\text k}$
2. $-8\hat{\text i} - 8\hat{\text j} - 6\hat{\text k}$
3. $-6\hat{\text i} - 6\hat{\text j} - 8\hat{\text k}$
4. $8\hat{\text i} + 8\hat{\text j} - 6\hat{\text k}$

Answer

$-6\hat{\text i} - 6\hat{\text j} - 8\hat{\text k}$

Reason — Given, q = 1, $\vec{\text v} = 2\hat{\text i} + 4\hat{\text j} + 6\hat{\text k}$, $\vec{\text B} = \text B\hat{\text i} + \text B\hat{\text j} + \text B_0\hat{\text k}$ and $\vec{\text F} = 4\hat{\text i} - 20\hat{\text j} + 12\hat{\text k}$.

Evaluating the vector product,

$\vec{\text v} \times \vec{\text B} = \begin{vmatrix} \hat{\text i} & \hat{\text j} & \hat{\text k} \\ 2 & 4 & 6 \\ \text B & \text B & \text B_0 \end{vmatrix} \\[1em] = \hat{\text i}(4\text B_0 - 6\text B) - \hat{\text j}(2\text B_0 - 6\text B) + \hat{\text k}(2\text B - 4\text B)$

Comparing with $\vec{\text F} \space$, from the $\hat{\text k}$ component,

$-2\text B = 12 \\[1em] \text B = -6$

From the $\hat{\text i}$ component,

$4\text B_0 - 6(-6) = 4 \\[1em] 4\text B_0 = 4 - 36 \\[1em] 4\text B_0 = -32 \\[1em] \text B_0 = -8$

Checking with the $\hat{\text j}$ component, $-[2(-8) - 6(-6)] = -[-16 + 36] = -20$, which agrees.

$\vec{\text B} = -6\hat{\text i} - 6\hat{\text j} - 8\hat{\text k}$

There are two infinitely long straight current carrying conductors and they are held at right angles to each other so that their common ends meet at the origin as shown in the given figure. The ratio of current in both conductors is 1 : 1. The magnetic field at point P is :

1. $\dfrac{\text µ_0 \text I}{4\text{πxy}}[\sqrt{\text x^2 + \text y^2} + (\text x + \text y)]$
2. $\dfrac{\text µ_0 \text I}{4\text{πxy}}[\sqrt{\text x^2 + \text y^2} - (\text x + \text y)]$
3. $\dfrac{\text µ_0 \text{Ixy}}{4\text π}[\sqrt{\text x^2 + \text y^2} - (\text x + \text y)]$
4. $\dfrac{\text µ_0 \text{Ixy}}{4\text π}[\sqrt{\text x^2 + \text y^2} + (\text x + \text y)]$

Answer

$\dfrac{\text µ_0 \text I}{4\text{πxy}}[\sqrt{\text x^2 + \text y^2} + (\text x + \text y)]$

Reason — Given, two semi-infinite conductors along the X and the Y axes with their common end at the origin, carrying equal currents I, and P the point (x, y).

For a straight conductor of finite length,

$\text B = \dfrac{\text µ_0}{4\text π}\dfrac{\text I}{\text d}(\sin \text φ_1 + \sin \text φ_2)$

Due to the conductor along the X-axis : the perpendicular distance of P is y; the angles are 90° (end at infinity) and θ₁, with $\sin \text θ_1 = \dfrac{\text x}{\sqrt{\text x^2 + \text y^2}}$. Hence

$\text B_1 = \dfrac{\text µ_0 \text I}{4\text{πy}}\left[1 + \dfrac{\text x}{\sqrt{\text x^2 + \text y^2}}\right]$

Due to the conductor along the Y-axis : the perpendicular distance of P is x; the angles are 90° (end at infinity) and θ₂, with $\sin \text θ_2 = \dfrac{\text y}{\sqrt{\text x^2 + \text y^2}}$. Hence

$\text B_2 = \dfrac{\text µ_0 \text I}{4\text{πx}}\left[1 + \dfrac{\text y}{\sqrt{\text x^2 + \text y^2}}\right]$

Both the fields at P are in the same direction, so writing $\text R = \sqrt{\text x^2 + \text y^2}$ and adding,

$\text B = \dfrac{\text µ_0 \text I}{4\text{πxy}}\left[\text x + \dfrac{\text x^2}{\text R} + \text y + \dfrac{\text y^2}{\text R}\right] \\[1em] \text B = \dfrac{\text µ_0 \text I}{4\text{πxy}}\left[(\text x + \text y) + \dfrac{\text x^2 + \text y^2}{\text R}\right] \\[1em] \text B = \dfrac{\text µ_0 \text I}{4\text{πxy}}\left[\sqrt{\text x^2 + \text y^2} + (\text x + \text y)\right]$

A coil having N turns is wound tightly in the form of a spiral with inner and outer radii a and b, respectively. Find the magnetic field at centre, when a current I passes through coil.

1. $\dfrac{\text µ_0 \text{IN}}{2(\text b - \text a)}\ln\left(\dfrac{\text b}{\text a}\right)$
2. $\dfrac{\text µ_0 \text I}{8}\ln\left[\dfrac{\text a + \text b}{\text a - \text b}\right]$
3. $\dfrac{\text µ_0 \text I}{4(\text a - \text b)}\ln\left[\dfrac{1}{\text a} - \dfrac{1}{\text b}\right]$
4. $\dfrac{\text µ_0 \text I}{8}\ln\left(\dfrac{\text a - \text b}{\text a + \text b}\right)$

Answer

$\dfrac{\text µ_0 \text{IN}}{2(\text b - \text a)}\ln\left(\dfrac{\text b}{\text a}\right)$

Reason — Given, a spiral of N turns wound tightly between the inner radius a and the outer radius b, carrying a current I.

The N turns are spread uniformly over the radial width (b − a), so the number of turns lying between the radii r and r + dr is

$\text{dN} = \dfrac{\text N}{\text b - \text a}\text{dr}$

Each of these turns behaves as a circular coil of radius r, so the field it produces at the centre is

$\text{dB} = \dfrac{\text µ_0 \text I}{2\text r}\text{dN} \\[1em] \text{dB} = \dfrac{\text µ_0 \text{IN}}{2(\text b - \text a)}\dfrac{\text{dr}}{\text r}$

Integrating from r = a to r = b,

$\text B = \dfrac{\text µ_0 \text{IN}}{2(\text b - \text a)}\int_\text a^\text b \dfrac{\text{dr}}{\text r} \\[1em] \text B = \dfrac{\text µ_0 \text{IN}}{2(\text b - \text a)}\left[\ln \text r\right]_\text a^\text b \\[1em] \text B = \dfrac{\text µ_0 \text{IN}}{2(\text b - \text a)}(\ln \text b - \ln \text a) \\[1em] \text B = \dfrac{\text µ_0 \text{IN}}{2(\text b - \text a)}\ln\left(\dfrac{\text b}{\text a}\right)$

The fractional change in the magnetic field intensity at a distance r from centre on the axis of current carrying coil of radius a to the magnetic field intensity at the centre of the same coil is : (Take, r < a)

1. $\dfrac{3}{2}\dfrac{\text a^2}{\text r^2}$
2. $\dfrac{2}{3}\dfrac{\text a^2}{\text r^2}$
3. $\dfrac{2}{3}\dfrac{\text r^2}{\text a^2}$
4. $\dfrac{3}{2}\dfrac{\text r^2}{\text a^2}$

Answer

$\dfrac{3}{2}\dfrac{\text r^2}{\text a^2}$

Reason — Given, a coil of radius a, and a point on its axis at a distance r from the centre, with r < a.

At the centre :