Moving Charges and Magnetic Field — Hit the Target

An electron is moving vertically upwards in the uniform magnetic field of the earth. The velocity of the electron is 2.0 × 10⁶ ms^-1 and the horizontal component of the earth magnetic field is 0.32 × 10^-4 T. Obtain the magnitude and direction of the force on the direction.

Answer

Given,

• v = 2.0 × 10⁶ ms^-1, vertically upwards
• B_H = 0.32 × 10^-4 T, horizontal and towards the north
• e = 1.6 × 10^-19 C

The electron moves vertically while the horizontal component of the earth's field is towards the north, so θ = 90° and

$\text F = \text{B}_\text H \text{ev} \\[1em] \text F = (0.32 \times 10^{-4})(1.6 \times 10^{-19})(2.0 \times 10^6) \\[1em] \text F = 1.024 \times 10^{-17}\ \text N$

The electron is negatively charged, so the direction of current is vertically downwards. Applying Fleming's left-hand rule with the current directed downwards and the field towards the north, the force on the electron is directed towards the east.

Hence, the force on the electron is 1.024 × 10^-17 N, directed towards the east.

Two very long conductors carrying currents I₁ and I₂ are separated by a distance d as shown in the figure. Give the expressions (do not derive) for magnetic field $\vec{\text B}_{12} \space$and $\vec{\text B}_{21} \space$at P and Q, respectively, produced by the electric currents I₂ and I₁. What are the directions of these fields? Define the S.I. Unit of current using the above set up.

Answer

The magnetic field at P, produced by the current I₂ in the conductor 2, is

$\text B_{12} = \dfrac{\text µ_0}{2\text π}\dfrac{\text I_2}{\text d}$

and the magnetic field at Q, produced by the current I₁ in the conductor 1, is

$\text B_{21} = \dfrac{\text µ_0}{2\text π}\dfrac{\text I_1}{\text d}$

Directions : By the right-hand thumb rule, both the conductors carrying currents in the same direction, $\vec{\text B}_{12} \space$at P is perpendicular to the plane of the paper directed outwards, while $\vec{\text B}_{21} \space$at Q is perpendicular to the plane of the paper directed inwards.

Definition of the SI unit of current : Each conductor lies in the field of the other and therefore experiences a force. The force per unit length is

$\dfrac{\text F}{\text L} = \dfrac{\text µ_0}{2\text π}\dfrac{\text I_1 \text I_2}{\text d}$

Putting I₁ = I₂ = 1 A and d = 1 m,

$\dfrac{\text F}{\text L} = \dfrac{4\text π \times 10^{-7}}{2\text π} \\[1em] \dfrac{\text F}{\text L} = 2 \times 10^{-7}\ \text{N m}^{-1}$

Hence, one ampere is that steady current which, when maintained in each of two infinitely long, straight and parallel conductors of negligible cross-section placed in vacuum at a distance of 1 metre from each other, produces a force of 2 × 10^-7 newton per metre length on each conductor.

Show that the radius of the path followed by a charged particle entering perpendicularly in a magnetic field is directly proportional to its momentum.

Answer

Let a particle of mass m and charge q enter a uniform magnetic field $\vec{\text B} \space$with a velocity $\vec{\text v} \space$directed perpendicular to the field.

Since $\vec{\text v} \space$is perpendicular to $\vec{\text B} \space$, the particle is acted upon by a force of magnitude qvB which lies in the plane of the path and is always perpendicular to the velocity. Such a force does not change the magnitude of the velocity, it changes only its direction. Hence the particle moves along a circle of radius r, the magnetic force acting as the centripetal force,

$\text{qvB} = \dfrac{\text{mv}^2}{\text r}$

Solving for r,

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

But mv is the momentum p of the particle, so

$\text r = \dfrac{\text p}{\text{qB}}$

For a given particle in a given field, q and B are constants, so

$\text r \propto \text p$

Hence, the radius of the path followed by a charged particle entering perpendicularly in a magnetic field is directly proportional to its momentum.

A wire of length 10 cm carrying 10 A of current is placed at an angle of 53° with respect to a magnetic field $\vec{\text B} \space$of strength 0.1 tesla. Find the force experienced by the wire. [use : sin 53° = 4/5]

Answer

Given,

• l = 10 cm = 0.1 m
• I = 10 A
• θ = 53°, with sin 53° = 4/5
• B = 0.1 tesla

The force experienced by a current-carrying conductor in a magnetic field is

$\text F = \text{IB}l \sin \text θ$

Substituting the values,

$\text F = (10)(0.1)(0.1)\left(\dfrac{4}{5}\right) \\[1em] \text F = 0.1 \times 0.8 \\[1em] \text F = 0.08\ \text N$

Hence, the force experienced by the wire is 0.08 N.

A current I = 1.5 A is flowing through a long solenoid of 500 turns. What is the magnetic force on an electron moving with a velocity of 2.0 × 10⁶ ms^-1, along the axis of the solenoid, when it is the middle.

Answer

Given,

• I = 1.5 A in a long solenoid of 500 turns
• v = 2.0 × 10⁶ ms^-1, along the axis of the solenoid

Inside a long solenoid the magnetic field is uniform and is directed along the axis of the solenoid. The electron is moving along this very axis, so the angle between $\vec{\text v} \space$and $\vec{\text B} \space$is zero.

$\text F = \text{Bev} \sin \text θ \\[1em] \text F = \text{Bev} \sin 0° \\[1em] \text F = 0$

Hence, the magnetic force on the electron is zero.

In the figure, PQ is a long straight wire carrying a current of 5A in the direction of the arrow. An electron travels with a velocity of 10⁶ m/s at a distance of 10 cm from the wire in the same direction of the current. Calculate the magnitude and direction of the force experienced by the electron.

Answer

Given,

• I = 5 A in the wire PQ
• v = 10⁶ m/s, in the same direction as the current
• r = 10 cm = 0.1 m

The magnetic field produced by the wire at the position of the electron is

$\text B = \dfrac{\text µ_0}{4\text π}\dfrac{2\text I}{\text r} \\[1em] \text B = (10^{-7}) \times \dfrac{2 \times 5}{0.1} \\[1em] \text B = 1 \times 10^{-5}\ \text T$

This field is perpendicular to the plane of the paper, so the electron moves at right angles to it and

$\text F = \text{evB} \\[1em] \text F = (1.6 \times 10^{-19})(10^6)(1 \times 10^{-5}) \\[1em] \text F = 1.6 \times 10^{-18}\ \text N$

The electron is negatively charged, so the direction of current is opposite to its motion. Applying Fleming's left-hand rule, the force on the electron is directed towards the right, that is, away from the wire.

Hence, the force experienced by the electron is 1.6 × 10^-18 N, towards right (away from the wire).

State Ampere's circuital law.

Answer

Ampere's circuital law : The line integral of the magnetic field (magnetic flux density) $\vec{\text B} \space$around any closed path in free space is equal to µ₀ times the net current I threading through the area enclosed by the path,

$\oint \vec{\text B} \cdot \vec{\text{dl}} = \text µ_0 \text I$

where µ₀ is the permeability of free space and I is the steady current that does not change with time. The law holds for any assembly of currents and for any closed path.

(i) Write an expression of Biot-Savart's law in the vector form. Write an expression for the magnetic field at the centre of a circular coil of radius R, with N turns and carrying a current I.

(ii) A helium nucleus completes one round of a circle of radius 0.8 m in 2 s. Find the magnetic field at the centre of the circle.

Answer

(i) In vector form, Biot-Savart's law states that the magnetic field produced at a point by a current-element $\text I \vec{\text{dl}} \space$is

$\text d\vec{\text B} = \dfrac{\text µ_0}{4\text π}\dfrac{\text I(\vec{\text{dl}} \times \vec{\text r})}{\text r^3}$

where $\vec{\text r} \space$is the vector drawn from the current-element to the point.

The magnetic field at the centre of a circular coil of radius R having N turns and carrying a current I is

$\text B = \dfrac{\text µ_0 \text N \text I}{2\text R}$

(ii) Given,

• R = 0.8 m
• time for one round, T = 2 s
• charge of the helium nucleus, q = 2e = 3.2 × 10^-19 C

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

$\text I = \dfrac{\text q}{\text T} \\[1em] \text I = \dfrac{3.2 \times 10^{-19}}{2} \\[1em] \text I = 1.6 \times 10^{-19}\ \text A$

The magnetic field at the centre of the circle is

$\text B = \dfrac{\text µ_0 \text I}{2\text R} \\[1em] \text B = \dfrac{(4\text π \times 10^{-7})(1.6 \times 10^{-19})}{2 \times 0.8} \\[1em] \text B = \dfrac{2.01 \times 10^{-25}}{1.6} \\[1em] \text B = 12.56 \times 10^{-26}\ \text T$

Hence, the magnetic field at the centre of the circle is 12.56 × 10^-26 T.

Write, in vector form, the defining equation for a magnetic field B at a point.

Answer

$\vec{\text F} = \text q(\vec{\text v} \times \vec{\text B})$

Explanation:
The magnetic field B at a point is defined through the force experienced by a charge moving at that point, where q is the charge and $\vec{\text v} \space$its velocity. The magnitude of this force is

$\text F = \text{Bqv} \sin \text θ \\[1em] \text B = \dfrac{\text F}{\text{qv} \sin \text θ}$

θ being the angle between $\vec{\text v} \space$and $\vec{\text B} \space$.

An electron beam passes through crossed electric and magnetic fields of 3.4 × 10⁴ Vm^-1 and 2 × 10^-3 Wbm^-2, respectively. If the path of the beam remains undeviated, calculate the speed of the electrons.

If the electric field is removed, what will be the radius of the circular path described by the electrons?

Answer

Given,

• E = 3.4 × 10⁴ Vm^-1
• B = 2 × 10^-3 Wbm^-2
• m_e = 9.1 × 10^-31 kg, e = 1.6 × 10^-19 C

Speed of the electrons : The path of the beam remains undeviated, so the electric force balances the magnetic force,

$\text{eE} = \text{evB} \\[1em] \text v = \dfrac{\text E}{\text B} \\[1em] \text v = \dfrac{3.4 \times 10^4}{2 \times 10^{-3}} \\[1em] \text v = 1.7 \times 10^7\ \text{m/s}$

Radius of the circular path : When the electric field is removed, only the magnetic force acts and the electrons describe a circle of radius

$\text r = \dfrac{\text{mv}}{\text{eB}} \\[1em] \text r = \dfrac{(9.1 \times 10^{-31})(1.7 \times 10^7)}{(1.6 \times 10^{-19})(2 \times 10^{-3})} \\[1em] \text r = \dfrac{1.547 \times 10^{-23}}{3.2 \times 10^{-22}} \\[1em] \text r = 4.83 \times 10^{-2}\ \text m$

Hence, the speed of the electrons is 1.7 × 10⁷ m/s and the radius of the circular path is 4.83 × 10^-2 m.

Write an equation of Lorentz force $\vec{\text F} \space$acting on a charged particle having charge q moving in a magnetic field $\vec{\text B} \space$in vector form.

Answer

$\vec{\text F} = \text q(\vec{\text v} \times \vec{\text B})$

Explanation:
This is the Lorentz force acting on a charged particle having charge q moving with velocity $\vec{\text v} \space$in a magnetic field $\vec{\text B} \space$. Its magnitude is F = Bqv sin θ, where θ is the angle between $\vec{\text v} \space$and $\vec{\text B} \space$, and its direction is perpendicular to the plane containing $\vec{\text v} \space$and $\vec{\text B} \space$.

When an electric field $\vec{\text E} \space$is also present, the complete Lorentz force is

$\vec{\text F} = \text q(\vec{\text E} + \vec{\text v} \times \vec{\text B})$

State Biot-Savart's law. Derive an expression for the magnetic field at the centre of a circular current carrying coil of radius r, with N turn.

Answer

Biot-Savart's law : The magnetic field dB developed at a point due to a small current-element I dl is directly proportional to the current I, to the length dl of the element and to sin θ, where θ is the angle between the element and the line joining it to the point, and inversely proportional to the square of the distance r between them,

$\text{dB} = \dfrac{\text µ_0}{4\text π}\dfrac{\text I\ \text{dl} \sin \text θ}{\text r^2}$

Magnetic field at the centre of a circular coil :

Let a circular coil of radius r carry a current I, and let O be its centre. Consider a small current-element dl of the coil. Every element of a circle is perpendicular to the radius drawn to the centre, so θ = 90° and the field at O due to this element is

$\text{dB} = \dfrac{\text µ_0}{4\text π}\dfrac{\text I\ \text{dl} \sin 90°}{\text r^2} \\[1em] \text{dB} = \dfrac{\text µ_0}{4\text π}\dfrac{\text I\ \text{dl}}{\text r^2}$

The direction of the field is the same for all the elements of the coil, so the fields simply add up. Hence the field at O due to the entire coil is

$\text B = \dfrac{\text µ_0}{4\text π}\dfrac{\text I}{\text r^2}\sum \text{dl}$

But $\sum \text{dl} = 2\text{πr}$, the circumference of the coil, so

$\text B = \dfrac{\text µ_0}{4\text π}\dfrac{\text I}{\text r^2}(2\text{πr}) \\[1em] \text B = \dfrac{\text µ_0 \text I}{2\text r}$

If the coil has N turns, each turn contributes equally, so

$\text B = \dfrac{\text µ_0 \text N \text I}{2\text r}$

How should a current-carrying conductor be placed in a magnetic field with respect to the field, so that the force acting upon the conductor is maximum?

Answer

The conductor should be placed perpendicular to the field.

Explanation:
The force acting on a current-carrying conductor placed in a magnetic field is

$\text F = \text{IB}l \sin \text θ$

which is maximum when sin θ = 1, that is when θ = 90°. Then F_max = IBl.

When a charged particle is projected perpendicular to a uniform magnetic field, it described a circular path in which:

1. its speed remains constant
2. its velocity remains constant
3. its momentum remains constant
4. its kinetic energy increases

Answer

its speed remains constant

Reason — The magnetic force on the particle is always perpendicular to its direction of motion, so it does no work on the particle. Hence the speed, and with it the kinetic energy, remains constant.

Using Ampere's circuital law show that the magnetic flux density B at a point P at perpendicular distance 'a' from a long current carrying conductor is given by,

$\text B = \left(\dfrac{\text µ_0}{4\text π}\right)\dfrac{2\text I}{\text a}.$

(Statement of law not required.)

Answer

Let the long conductor carry a steady current I, and let P be a point at a perpendicular distance a from it.

Draw a circle of radius a through P, with its centre on the conductor and its plane perpendicular to the conductor. By symmetry, the magnitude of $\vec{\text B} \space$is the same at all points of this circle and its direction is tangential to the circle at each point. Hence $\vec{\text B} \space$and $\vec{\text{dl}} \space$are along the same direction at every element, and

$\oint \vec{\text B} \cdot \vec{\text{dl}} = \oint \text B\ \text{dl} \cos 0° \\[1em] \oint \vec{\text B} \cdot \vec{\text{dl}} = \text B \oint \text{dl} \\[1em] \oint \vec{\text B} \cdot \vec{\text{dl}} = \text B(2\text{πa})$

The current enclosed by the circle is I, so applying Ampere's circuital law,

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

Hence, the magnetic flux density at P is $\text B = \left(\dfrac{\text µ_0}{4\text π}\right)\dfrac{2\text I}{\text a}$.

How much force per unit length acts on a long current carrying conductor X due to a current flowing through another similar conductor Y, kept parallel to it in vacuum?

Answer

$\dfrac{\text F}{\text L} = \dfrac{\text µ_0}{2\text π}\dfrac{\text I_1 \text I_2}{\text r}$

Explanation:
Each conductor lies in the magnetic field produced by the current in the other, and so experiences a force.

If the conductors X and Y carry currents I₁ and I₂ and are kept parallel at a distance r apart in vacuum, the field produced by Y at the position of X is

$\text B = \dfrac{\text µ_0}{2\text π}\dfrac{\text I_2}{\text r}$

The conductor X lies at right angles to this field, so the force on a length L of it is F = I₁BL. Dividing by L gives the force per unit length stated above.

This force is one of attraction if the currents are in the same direction, and one of repulsion if they are in opposite directions.

Electrons having a velocity $\vec{\text v} \space$of 2 × 10⁶ ms^-1 pass undeviated through a uniform electric field $\vec{\text E} \space$of intensity 5 × 10⁴ Vm^-1 and a uniform magnetic field $\vec{\text B} \space$.

(i) Find the magnitude of magnetic flux density B of the magnetic field.

(ii) What is the direction of $\vec{\text B} \space$, if $\vec{\text v} \space$is towards right and $\vec{\text E} \space$is vertically downwards in the plane of this paper?

Answer

Given,

• v = 2 × 10⁶ ms^-1
• E = 5 × 10⁴ Vm^-1

(i) The electrons pass undeviated, so the electric force balances the magnetic force,

$\text{eE} = \text{evB} \\[1em] \text B = \dfrac{\text E}{\text v} \\[1em] \text B = \dfrac{5 \times 10^4}{2 \times 10^6} \\[1em] \text B = 2.5 \times 10^{-2}\ \text T$

(ii) The three vectors $\vec{\text v} \space$, $\vec{\text E} \space$and $\vec{\text B} \space$are mutually perpendicular. With $\vec{\text v} \space$towards the right and $\vec{\text E} \space$vertically downwards in the plane of the paper, $\vec{\text B} \space$must be perpendicular to the plane of the paper. For the two forces on the electron to be opposite, the field must be directed into the plane of the paper (inwards).

Hence, the magnetic flux density is 2.5 × 10^-2 T, directed perpendicular to the plane of paper and inwards.

Figure shows a point P near a long conductor XY carrying a current I, MN is a short current-carrying conductor, kept at the point P, parallel to the conductor XY.

(i) What is the direction of magnetic flux density 'B' at the point P due to the current flowing through XY?

(ii) What is the direction of the force experienced by the conductor MN due to the current flowing through XY?

Answer

(i) The current I in the conductor XY flows upwards, and the point P lies to the right of XY. Applying the right-hand thumb rule, the magnetic flux density B at P is perpendicular to the plane of the paper, directed into the page.

(ii) The short conductor MN at P carries a current I in the same direction as the current in XY, so the two currents are parallel. Parallel currents attract each other, so the force experienced by MN is directed towards the conductor XY.

The same result follows from Fleming's left-hand rule, with the field at P into the page and the current in MN directed upwards.

Figure shows two infinitely long and thin current-carrying conductors X and Y kept in vacuum, parallel to each other, at a distance 'a'.

(i) How much force per unit length acts on the conductor Y due to the current flowing through X? Write your answer in terms of $\left(\dfrac{\text µ_0}{4\text π}\right)$, I₁, I₂ and a. (Derivation of formula is not required).

(ii) Define one ampere, in terms of force between two current-carrying conductors.

Answer

(i) The field produced by the conductor X at the position of the conductor Y is

$\text B = \left(\dfrac{\text µ_0}{4\text π}\right)\dfrac{2\text I_1}{\text a}$

The conductor Y lies at right angles to this field, so the force on a length L of Y is F = I₂BL, and the force per unit length is

$\dfrac{\text F}{\text L} = \left(\dfrac{\text µ_0}{4\text π}\right)\dfrac{2\text I_1 \text I_2}{\text a}$

(ii) Definition of one ampere : One ampere is that steady current which, when maintained in each of two infinitely long, straight and parallel conductors of negligible cross-section placed in vacuum at a distance of 1 metre from each other, produces a force of 2 × 10^-7 newton per metre length on each conductor.

The Biot-Savart's law in vector form is :

1. $\delta\vec{\text B} = \dfrac{\text µ_0}{4\text π}\dfrac{\text{dl}(\vec{\text I} \times \vec{\text r})}{\text r^3}$
2. $\delta\vec{\text B} = \dfrac{\text µ_0}{4\text π}\dfrac{\text I(\vec{\text{dl}} \times \vec{\text r})}{\text r^3}$
3. $\delta\vec{\text B} = \dfrac{\text µ_0}{4\text π}\dfrac{\text I(\vec{\text r} \times \vec{\text{dl}})}{\text r^3}$
4. $\delta\vec{\text B} = \dfrac{\text µ_0}{4\text π}\dfrac{\text I(\vec{\text{dl}} \times \vec{\text r})}{\text r^2}$

Answer

$\delta\vec{\text B} = \dfrac{\text µ_0}{4\text π}\dfrac{\text I(\vec{\text{dl}} \times \vec{\text r})}{\text r^3}$

Reason — In vector form Biot-Savart's law is written with the current-element as the vector $\text I \vec{\text{dl}}$, and the field is perpendicular to the plane containing $\vec{\text{dl}} \space$and $\vec{\text r} \space$. Writing $\vec{\text r} = \text r\hat{\text r}$, the scalar form

$\delta \text B = \dfrac{\text µ_0}{4\text π}\dfrac{\text I\ \text{dl} \sin \text θ}{\text r^2}$

becomes $\delta\vec{\text B} = \dfrac{\text µ_0}{4\text π}\dfrac{\text I(\vec{\text{dl}} \times \vec{\text r})}{\text r^3}$, with r³ in the denominator because the numerator carries one extra factor of r.

The particles which cannot be accelerated by a cyclotron or a van de Graaff generator are :

1. alpha particles
2. beta particles
3. neutrons
4. protons

Answer

neutrons

Reason — Both a cyclotron and a van de Graaff generator accelerate a particle by means of an electric field, and the force exerted by an electric field on a particle is F = qE. A neutron carries no charge, so no electric force acts on it and it cannot be accelerated by either machine.

A metallic wire having length of 2 m and weight of 4 × 10^-3 N is found to remain at rest in a uniform and transverse magnetic field of 2 × 10^-4 T. Current flowing through the wire is :

1. 10 A
2. 5 A
3. 2 A
4. 1 A

Answer

10 A

Reason — Given, l = 2 m, weight W = 4 × 10^-3 N and B = 2 × 10^-4 T.

The wire remains at rest, so the magnetic force balances its weight,

$\text{BI}l = \text W \\[1em] \text I = \dfrac{\text W}{\text B l} \\[1em] \text I = \dfrac{4 \times 10^{-3}}{(2 \times 10^{-4})(2)} \\[1em] \text I = \dfrac{4 \times 10^{-3}}{4 \times 10^{-4}} \\[1em] \text I = 10\ \text A$

When does a moving charged particle not experience any force while moving through a uniform magnetic field?

Answer

A moving charged particle experiences no force while moving through a uniform magnetic field when it moves parallel or anti-parallel to the direction of the field.

Explanation:
The magnetic force on a charged particle is

$\text F = \text{Bqv} \sin \text θ$

where θ is the angle between $\vec{\text v} \space$and $\vec{\text B} \space$. This force is zero when sin θ = 0, that is when θ = 0° or 180°.

A circular coil carrying a current I has radius R and number of turns N. If all the three, i.e., the current I, radius R and number of turns N are doubled, then magnetic field at its centre becomes :

1. double
2. half
3. four times
4. one-fourth

Answer

double

Reason — The magnetic field at the centre of a circular coil is

$\text B = \dfrac{\text µ_0 \text N \text I}{2\text R}$

When the current, the radius and the number of turns are all doubled, that is I → 2I, R → 2R and N → 2N,

$\text B' = \dfrac{\text µ_0 (2\text N)(2\text I)}{2(2\text R)} \\[1em] \text B' = 2 \times \dfrac{\text µ_0 \text N \text I}{2\text R} \\[1em] \text B' = 2\text B$

A moving charged particle q travelling along the positive x-axis enters a uniform magnetic field B. When will the force acting on q be maximum?

Answer

The force acting on q will be maximum when it moves perpendicular to the magnetic field B, that is, when θ = 90°.

Explanation:
The magnetic force on the charge is

$\text F = \text{Bqv} \sin \text θ$

which is maximum when sin θ = 1, that is θ = 90°, giving F_max = Bqv. As the particle travels along the positive x-axis, the force is maximum when the field B is directed at right angles to the x-axis.

A long horizontal wire P carries a current of 50A. It is rigidly fixed. Another wire Q is placed directly above the parallel to P, as shown in figure.

The weight per unit length of the wire Q is 0.025 Nm^-1 and it carries a current of 25A. Find the distance 'r' of the wire Q from the wire P so that the wire Q remains at rest.

Answer

Given,

• I₁ = 50 A in the wire P
• I₂ = 25 A in the wire Q
• weight per unit length of Q = 0.025 Nm^-1

The wire Q remains at rest above P, so the magnetic force per unit length on it must be directed upwards and must balance its weight per unit length. From the figure the two currents are in opposite directions, so the wires repel each other and Q is supported.

$\dfrac{\text µ_0}{2\text π}\dfrac{\text I_1 \text I_2}{\text r} = 0.025 \\[1em] (2 \times 10^{-7}) \times \dfrac{(50)(25)}{\text r} = 0.025 \\[1em] \dfrac{2.5 \times 10^{-4}}{\text r} = 0.025 \\[1em] \text r = \dfrac{2.5 \times 10^{-4}}{0.025} \\[1em] \text r = 0.01\ \text m = 1\ \text{cm}$

Hence, the wire Q should be kept at a distance of 1 cm from the wire P.

Calculate force per unit length acting on the wire B due to the current flowing in the wire A. (Shown in the figure).

Answer

Given,

• I₁ = 75 A in the wire A
• I₂ = 20 A in the wire B
• r = 1 cm = 0.01 m

The force per unit length acting on the wire B due to the current in the wire A is

$\dfrac{\text F}{\text L} = \dfrac{\text µ_0}{2\text π}\dfrac{\text I_1 \text I_2}{\text r}$

Substituting the values,

$\dfrac{\text F}{\text L} = (2 \times 10^{-7}) \times \dfrac{(75)(20)}{0.01} \\[1em] \dfrac{\text F}{\text L} = (2 \times 10^{-7})(1.5 \times 10^5) \\[1em] \dfrac{\text F}{\text L} = 3 \times 10^{-2}\ \text{N/m}$

The currents in the two wires are in opposite directions, so the force is one of repulsion, directed away from the wire A.

Hence, the force per unit length acting on the wire B is 3 × 10^-2 N/m.

(a) An electric current flows through an infinitely long conductor as shown in figure. Write an expression and direction for the magnetic field at point P.

(b) An electric current I flows through a circular loop as shown in figure. Write an expression and direction for the magnetic field at the centre of the loop at point?

Answer

(a)

For an infinitely long straight conductor carrying a current I, the magnetic field at a point P at a perpendicular distance r from it is

$\text B = \dfrac{\text µ_0}{2\text π}\dfrac{\text I}{\text r}$

The current flows upwards and P lies to the right of the conductor, so by the right-hand thumb rule the field at P is perpendicular to the plane of the paper, directed into the page.

(b)

For a circular loop of radius R carrying a current I, the magnetic field at the centre is

$\text B = \dfrac{\text µ_0 \text I}{2\text R}$

The current flows in the anticlockwise sense, so by the right-hand palm rule no. 1 the field at the centre is perpendicular to the plane of the loop, directed out of the page (towards the observer).

Write Biot-Savart's law in vector form.

Answer

$\text d\vec{\text B} = \dfrac{\text µ_0}{4\text π}\dfrac{\text I(\vec{\text{dl}} \times \vec{\text r})}{\text r^3}$

Explanation:
Here $\text d\vec{\text B} \space$is the magnetic field produced at a point by a current-element $\text I \vec{\text{dl}}$, $\vec{\text r} \space$is the vector drawn from the current-element to the point at which the field is required, and µ₀ is the permeability of free space. The direction of $\text d\vec{\text B} \space$is perpendicular to the plane containing $\vec{\text{dl}} \space$and $\vec{\text r} \space$.

Write an expression for force per unit length between two long current carrying wires, kept parallel to each other in vacuum and hence define an ampere, the SI unit of current.

Answer

If two long conductors carrying currents I₁ and I₂ are kept parallel to each other at a distance r apart in vacuum, the force per unit length between them is

$\dfrac{\text F}{\text L} = \dfrac{\text µ_0}{2\text π}\dfrac{\text I_1 \text I_2}{\text r}$

This force is attractive if the currents are in the same direction and repulsive if they are in opposite directions.

Definition of ampere : Putting I₁ = I₂ = 1 A and r = 1 m,

$\dfrac{\text F}{\text L} = \dfrac{4\text π \times 10^{-7}}{2\text π} \times \dfrac{1 \times 1}{1} \\[1em] \dfrac{\text F}{\text L} = 2 \times 10^{-7}\ \text{N m}^{-1}$

Hence, one ampere is that steady current which, when maintained in each of two infinitely long, straight and parallel conductors of negligible cross-section placed in vacuum at a distance of 1 metre from each other, produces a force of 2 × 10^-7 newton per metre length on each conductor.

Magnetic flux density or the magnetic field, at a point on the axis of a long solenoid is given by :

1. B = µ₀nI
2. $\text B = \dfrac{\text µ_0 \text I}{2\text r}$
3. $\text B = \dfrac{\text µ_0 \text I}{4\text{πr}}$
4. B = µ₀I

Answer

B = µ₀nI

Reason — Applying Ampere's circuital law to a rectangular path partly inside and partly outside a long solenoid gives Bx = µ₀(nx)I, so

$\text B = \text µ_0 \text n \text I$

where n is the number of turns per unit length.

Write an equation of Lorentz force in vector form.

Answer

$\vec{\text F} = \text q(\vec{\text E} + \vec{\text v} \times \vec{\text B})$

Explanation:
The Lorentz force is the total force acting on a charged particle moving in a region where both an electric field $\vec{\text E} \space$and a magnetic field $\vec{\text B} \space$are present. Here q is the charge on the particle and $\vec{\text v} \space$its velocity. The first term is the electric force, which acts whether the charge is at rest or in motion, and the second term is the magnetic force, which acts only on a moving charge.

A proton is moving in a uniform magnetic field B with a velocity v at an angle θ with the field. The force acting on the proton is minimum when θ is :

1. 120°
2. 90°
3. 60°
4. 0°

Answer

0°

Reason — The force on the proton is

$\text F = \text{Bqv} \sin \text θ$

which is minimum when sin θ is minimum. Among the given angles sin θ is least at θ = 0°, where sin 0° = 0 and the force becomes zero. At θ = 90° the force is maximum.

Two long straight wires P and Q are parallel to each other. They carry currents in the same direction as shown in the figure.

The wire P is kept fixed. Then, the wire Q will :

1. also remain at rest
2. move towards the wire P
3. move away from the wire P
4. come out of the plane of this paper

Answer

move towards the wire P

Reason — The two wires carry currents in the same direction, so they attract each other. The force per unit length between them is

$\dfrac{\text F}{\text L} = \dfrac{\text µ_0}{2\text π}\dfrac{\text I_1 \text I_2}{\text r}$

directed towards the other wire. As the wire P is kept fixed, the wire Q moves towards P.

A circular coil has radius 'r', number of turns 'N' and carries a current 'I'. Magnetic flux density 'B' at its centre is :

1. B = µ₀NI
2. B = µ₀NI/2r
3. B = µ₀NI/4πr
4. B = µ₀NI/4r

Answer

B = µ₀NI/2r

Reason — Applying Biot-Savart's law to every element of the coil and adding the fields, which are all in the same direction at the centre,

$\text B = \dfrac{\text µ_0}{4\text π}\dfrac{\text I}{\text r^2}(2\text{πr}) \\[1em] \text B = \dfrac{\text µ_0 \text I}{2\text r}$

for a single turn. For N turns each turn contributes equally, so $\text B = \dfrac{\text µ_0 \text{NI}}{2\text r}$.

A long straight wire AB carries a current of 5 A. P is a proton travelling with a velocity of 2 × 10⁶ m/s, parallel to the wire, 0.2 m from it and in a direction opposite to the current, as shown in the given figure. Calculate the force which magnetic field of the current-carrying conductor AB exerts on the proton.

Answer

Given,

• I = 5 A in the wire AB
• v = 2 × 10⁶ m/s, parallel to the wire but opposite to the current
• r = 0.2 m

The magnetic field produced by the conductor at the position of the proton is

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

This field is perpendicular to the plane of the paper, so the proton moves at right angles to it and

$\text F = \text{qvB} \\[1em] \text F = (1.6 \times 10^{-19})(2 \times 10^6)(5 \times 10^{-6}) \\[1em] \text F = 1.6 \times 10^{-18}\ \text N$

Applying Fleming's left-hand rule, this force is directed towards the right hand side, that is, away from the wire.

Hence, the force exerted on the proton is 1.6 × 10^-18 N, towards right hand side (away from the wire).

Using Ampere's circuital law, obtain an expression for magnetic flux density 'B' at a point near an infinitely long and straight conductor, carrying a current I.

Answer

Let an infinitely long straight conductor carry a steady current I, and let P be a point at a perpendicular distance r from it.

Draw a circle of radius r through P, with its centre on the conductor and its plane perpendicular to the conductor. By symmetry, the magnitude of $\vec{\text B} \space$is the same at all points of this circle and its direction is tangential to the circle at each point. Hence $\vec{\text B} \space$and $\vec{\text{dl}} \space$are along the same direction at every element of the circle, and

$\oint \vec{\text B} \cdot \vec{\text{dl}} = \oint \text B\ \text{dl} \cos 0° \\[1em] \oint \vec{\text B} \cdot \vec{\text{dl}} = \text B \oint \text{dl}$

But $\oint \text{dl} = 2\text{πr}$, the circumference of the circle, and the current enclosed by the circle is I. Applying Ampere's circuital law,

$\text B(2\text{πr}) = \text µ_0 \text I \\[1em] \text B = \dfrac{\text µ_0}{2\text π}\dfrac{\text I}{\text r}$

Hence, the magnetic flux density at a point near an infinitely long straight conductor is $\text B = \dfrac{\text µ_0}{2\text π}\dfrac{\text I}{\text r}$.

Biot Savart's law in vector form is : [K = µ₀/4π]

1. $\delta\vec{\text B} = \text{KI}{\vec{\text{dl}} \times \vec{\text r}}/\sqrt{\text r}$
2. $\delta\vec{\text B} = \text{KI}{\vec{\text{dl}} \times \vec{\text r}}/\text r$
3. $\delta\vec{\text B} = \text{KI}{\vec{\text{dl}} \times \vec{\text r}}/\text r^2$
4. $\delta\vec{\text B} = \text{KI}{\vec{\text{dl}} \times \vec{\text r}}/\text r^3$

Answer

$\delta\vec{\text B} = \text{KI}{\vec{\text{dl}} \times \vec{\text r}}/\text r^3$

Reason — Biot-Savart's law in scalar form is

$\delta \text B = \text{KI}\dfrac{\text{dl} \sin \text θ}{\text r^2}, \qquad \text K = \dfrac{\text µ_0}{4\text π}$

Since $|\vec{\text{dl}} \times \vec{\text r}| = \text{dl}\ \text r \sin \text θ$, writing the numerator as a cross product introduces one extra factor of r, which must be compensated by r³ in the denominator.

A straight metallic wire of length of 20 cm weights 1.2g. It is kept perpendicular to a uniform magnetic field of 0.3 T. How much current should be passed through it so that it does not fall due to the action of gravity?

Answer

Given,

• l = 20 cm = 0.2 m
• m = 1.2 g = 1.2 × 10^-3 kg
• B = 0.3 T
• g = 9.8 m s^-2

The wire does not fall, so the magnetic force on it balances its weight,

$\text{BI}l = \text{mg} \\[1em] \text I = \dfrac{\text{mg}}{\text B l} \\[1em] \text I = \dfrac{(1.2 \times 10^{-3})(9.8)}{(0.3)(0.2)} \\[1em] \text I = \dfrac{11.76 \times 10^{-3}}{0.06} \\[1em] \text I = 0.196\ \text A$

Hence, a current of 0.196 A (that is, 196 × 10^-3 A) should be passed through the wire.

Note: The printed answer key gives 19.6 × 10^-3 A. Substituting the given data in BIl = mg yields 196 × 10^-3 A, so the printed value appears to be smaller by a factor of ten.

Lorentz force in vector form is :

1. F = Bqv sin θ
2. $\vec{\text F} = \text q(\vec{\text v} \times \vec{\text B})$
3. $\vec{\text F} = \text q(\vec{\text B} \times \vec{\text v})$
4. $\vec{\text F} = \vec{\text v}(\text q \times \vec{\text B})$

Answer

$\vec{\text F} = \text q(\vec{\text v} \times \vec{\text B})$

Reason — The Lorentz force on a charge q moving with velocity $\vec{\text v} \space$in a magnetic field $\vec{\text B} \space$is the vector product of $\vec{\text v} \space$and $\vec{\text B} \space$multiplied by q, taken in that order.

Which one of the following copper wires has minimum resistance?

1. length = x, area of cross section = y
2. length = 2x, area of cross section = y
3. length = x, area of cross section = 2y
4. length = 2x, area of cross section = 2y

Answer

length = x, area of cross section = 2y

Reason — The resistance of a wire is

$\text R = \text ρ\dfrac{l}{\text A}$

so R is least when the length is least and the area of cross-section is greatest. Among the four wires, the one with length x and area 2y gives

$\text R = \dfrac{\text ρ \text x}{2\text y}$

which is the smallest of the four values $\dfrac{\text ρ \text x}{\text y}$, $\dfrac{2\text ρ \text x}{\text y}$, $\dfrac{\text ρ \text x}{2\text y}$ and $\dfrac{\text ρ \text x}{\text y}$.

Note: The printed answer key gives option 1 (length = x, area = y), but that wire has resistance $\dfrac{\text ρ \text x}{\text y}$, which is twice the resistance of the wire of option 3. Hence option 3 is the correct choice.

Figure below shows an infinitely long metallic wire YY' which is carrying a current I'. P is a point at a perpendicular distance r from it.

(i) What is the direction of magnetic flux density B of the magnetic field at the point P?

(ii) What is the magnitude of magnetic flux density B of the magnetic field at the point P?

(iii) Another metallic wire MN having length l and carrying a current I is now kept at the point P. If the two wires are in vacuum and parallel to each other, how much force acts on the wire MN due to the current I', flowing in the wire YY'?

Answer

(i) The current I' in the wire YY' flows upwards and the point P lies to the right of the wire. By the right-hand thumb rule, the magnetic flux density B at P is perpendicular to the plane of the paper and directed downwards (into the page).

(ii) For an infinitely long straight conductor, the magnitude of the magnetic flux density at a perpendicular distance r is

$\text B = \dfrac{\text µ_0}{2\text π}\dfrac{\text I'}{\text r}$

(iii) The wire MN of length l carrying a current I is placed at P, parallel to YY', and lies at right angles to the field B. Hence the force on it is

$\text F = \text{BI}l \\[1em] \text F = \dfrac{\text µ_0}{2\text π}\dfrac{\text I'\ \text I\ l}{\text r}$

This force is one of attraction if the two currents are in the same direction, and one of repulsion if they are in opposite directions.

Show that force per unit length (F') between two long wires carrying currents I and I' kept parallel to each other in vacuum, is given by :

$\text F' = \left(\dfrac{\text µ_0}{4\text π}\right)\dfrac{2\text{II}'}{\text r}$

where the terms have their usual meaning.

Answer

Let two long straight wires be kept parallel to each other in vacuum at a distance r apart, carrying currents I and I' respectively.

The magnitude of the magnetic field produced by the current I at the position of the second wire is

$\text B = \dfrac{\text µ_0}{2\text π}\dfrac{\text I}{\text r}$

and by the right-hand palm rule no. 1 this field is perpendicular to the plane containing the two wires.

The second wire, carrying the current I', is thus situated in this field at right angles to it, so the force on a length L of it is

$\text F = \text I' \text B \text L \sin 90° \\[1em] \text F = \text I'\left(\dfrac{\text µ_0}{2\text π}\dfrac{\text I}{\text r}\right)\text L$

Therefore the force per unit length is

$\text F' = \dfrac{\text F}{\text L} \\[1em] \text F' = \dfrac{\text µ_0}{2\text π}\dfrac{\text I\ \text I'}{\text r}$

Writing $\dfrac{\text µ_0}{2\text π} = \left(\dfrac{\text µ_0}{4\text π}\right)\times 2$,

$\text F' = \left(\dfrac{\text µ_0}{4\text π}\right)\dfrac{2\text{II}'}{\text r}$

Hence, the force per unit length between the two wires is $\text F' = \left(\dfrac{\text µ_0}{4\text π}\right)\dfrac{2\text{II}'}{\text r}$, where the terms have their usual meaning.