Moving Charges and Magnetism Chapter 4: NCERT Solutions

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NCERT Solutions for Class 12 Physics Chapter 4 Moving Charges and Magnetism

Question 1. A circular coil of wire consisting of \( 100 \) turns, each of radius \( 8.0 \, \text{cm} \) carries a current of \( 0.40 \, \text{A} \). What is the magnitude of the magnetic field \( B \) at the center of the coil?

Sol: The magnetic field \( B \) at the center of a circular coil is given by: \[ B = \frac{\mu_0}{4\pi} \cdot \frac{2\pi n I}{r} \] where: \begin{align*} n & = \text{number of turns} = 100 \\ I & = \text{current} = 0.40 \, \text{A} \\ r & = \text{radius of the coil} = 8.0 \, \text{cm} = 0.08 \, \text{m} \\ \mu_0 & = \text{permeability of free space} = 4\pi \times 10^{-7} \, \text{T m/A} \end{align*} Substitute these values into the formula: \[ B = \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{2\pi \times 100 \times 0.40}{0.08} \] \[ B = 10^{-7} \cdot \frac{2 \times 100 \times 0.40}{0.08} \] \[ B = 10^{-7} \cdot \frac{80}{0.08} \] \[ B = 10^{-7} \cdot 1000 \] \[ B = 3.1 \times 10^{-4} \, \text{T} \] Thus, the magnitude of the magnetic field \( B \) at the center of the coil is \( 3.1 \times 10^{-4} \, \text{T} \).

Question 2. A long straight wire carries a current of 35 A. What is the magnitude of the field B at a point 20 cm from the wire?

Solution :Given:

Current through the wire, I = 35 A

Distance of point P from the wire, d = 20 cm

Question 3. A long straight wire in the horizontal plane carries a current of 50 A in north to south direction. Give the magnitude and direction of B at a point 2.5 m east of the wire.

Solution :Given:

Current through the wire, I = 50 A (North to South)

Distance of point P East of the wire, d = 2.5 m

Direction of magnetic field,

The point is in a plane normal to the wire and the wire carries current in north to south. Using the Right hand thumb rule we can conclude that the direction of the magnetic field is vertically upwards, or out of the paper.

The magnitude of the magnetic field is 4 × 10-6T and its direction is upwards or out of paper.

Question 4. A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?

Solution :Given:

Current through the wire, I = 90 A (East to West)

Distance of point P below the wire, d = 1.5 m

Direction of magnetic field,

We know that wire carries current in the east to west direction. Using the Right hand thumb rule, we can conclude that the direction of the magnetic field is from north to south as indicated in the figure.

The magnitude of the magnetic field is 1.2 × 10-5T and its direction is from north to south.

Question5. What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30º with the direction of a uniform magnetic field of 0.15 T?

Solution :
Current in the wire, I = 8 A

Magnitude of the uniform magnetic field, B = 0.15 T

Angle between the wire and magnetic field, θ = 30°.

Magnetic force per unit length on the wire is given as:

f = BI sinθ

= 0.15 × 8 ×1 × sin 30°

= 0.6 N m–1

Hence, the magnetic force per unit length on the wire is 0.6 N m–1.

Question6. A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire?

Solution :
Length of the wire, l = 3 cm = 0.03 m

Current flowing in the wire, I = 10 A

Magnetic field, B = 0.27 T

Angle between the current and magnetic field, θ = 90°

Magnetic force exerted on the wire is given as:

F = BIlsinθ

= 0.27 × 10 × 0.03 sin 90°

= 8.1 × 10–2 N

Hence, the magnetic force on the wire is 8.1 × 10–2 N. The direction of the force can be obtained from Fleming’s left hand rule.

Question 7. Two long and parallel straight wires \( A \) and \( B \) carrying currents of \( 8.0 \, \text{A} \) and \( 5.0 \, \text{A} \) in the same direction are separated by a distance of \( 4.0 \, \text{cm} \). Estimate the force on a \( 10 \, \text{cm} \) section of wire \( A \).

Sol: The force per unit length \( \frac{F}{l} \) between two long parallel wires carrying currents \( i_1 \) and \( i_2 \) and separated by a distance \( x \) is given by: \[ \frac{F}{l} = \frac{\mu_0 i_1 i_2}{2 \pi x} \] where: \begin{align*} \mu_0 & = \text{permeability of free space} = 4\pi \times 10^{-7} \, \text{T m/A} \\ i_1 & = 8.0 \, \text{A} \\ i_2 & = 5.0 \, \text{A} \\ x & = 4.0 \, \text{cm} = 0.04 \, \text{m} \end{align*} Substitute these values into the formula: \[ \frac{F}{l} = \frac{4 \pi \times 10^{-7} \times 8.0 \times 5.0}{2 \pi \times 0.04} \] \[ \frac{F}{l} = \frac{4 \times 10^{-7} \times 40}{0.08} \] \[ \frac{F}{l} = \frac{1.6 \times 10^{-5}}{0.08} \] \[ \frac{F}{l} = 2 \times 10^{-4} \, \text{N/m} \] The length of wire \( A \) is \( 10 \, \text{cm} = 0.10 \, \text{m} \). So, the total force \( F \) on a \( 10 \, \text{cm} \) section of wire \( A \) is: \[ F = \frac{F}{l} \times l = 2 \times 10^{-4} \times 0.10 \] \[ F = 2 \times 10^{-5} \, \text{N} \] Thus, the force on a \( 10 \, \text{cm} \) section of wire \( A \) is \( 2 \times 10^{-5} \, \text{N} \).

Question8. A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near its centre.

Solution :Given:

Length of solenoid, L = 80cm

Number of turns = number of layers × number of turns per layer

Number of turns, n = 5 × 400 = 2000

Radius of solenoid, r = Diameter/2 = 0.9 cm

Current through the solenoid = 8.0A

Hence the magnetic field strength at the centre of the solenoid is 2.512 × 10-2T.

Question9. A square coil of side \( 10 \, \text{cm} \) consists of \( 20 \) turns and carries a current of \( 12 \, \text{A} \). The coil is suspended vertically and the normal to the plane of the coil makes an angle of \( 30^\circ \) with the direction of a uniform horizontal magnetic field of magnitude \( 0.80 \, \text{T} \). What is the magnitude of torque experienced by the coil?

Sol: The magnitude of the torque \( \tau \) experienced by the coil in the magnetic field is given by: \[ \tau = n B I A \sin \theta \] where: \begin{align*} n & = \text{number of turns} = 20 \\ B & = \text{magnetic field strength} = 0.80 \, \text{T} \\ I & = \text{current} = 12 \, \text{A} \\ A & = \text{area of the coil} \\ \theta & = \text{angle between the normal to the plane of the coil and the magnetic field} = 30^\circ \end{align*} The side of the square coil \( l = 10 \, \text{cm} = 0.10 \, \text{m} \). So, the area \( A \) of the coil is: \[ A = l \times l = (0.10 \, \text{m})^2 = 0.01 \, \text{m}^2 \] Substitute the values into the torque formula: \[ \tau = 20 \times 0.80 \times 12 \times 0.01 \times \sin 30^\circ \] \[ \tau = 20 \times 0.80 \times 12 \times 0.01 \times 0.5 \] \[ \tau = 0.96 \, \text{N m} \] Thus, the magnitude of torque experienced by the coil is \( 0.96 \, \text{N m} \).

Question10. Two moving coil meters, M1 and M2 have the following particulars:

R1 = 10 Ω, N1 = 30,

A1 = 3.6 × 10–3 m2, B1 = 0.25 T

R2 = 14 Ω, N2 = 42,

A2 = 1.8 × 10–3 m2, B2 = 0.50 T

(The spring constants are identical for the two meters).

Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M2 and M1.

Solution :Given:

For moving coil meter M1

Resistance of wire, R1 = 10Ω

Number of turns, N1 = 30

Area of cross-section, A1 = 3.6 × 10-3 m2

Magnetic field strength, B1 = 0.25 T

For moving coil meter M2

Resistance of wire, R2 = 14Ω

Number of turns, N2 = 42

Area of cross-section, A2 = 1.8 × 10-3 m2

Magnetic field strength, B2 = 0.50 T\

Spring constant, K1 = K2 = K

Current sensitivity is given by,

Hence, the ratio of current sensitivities is 1.4.

Hence, the ratio of voltage sensitivity of M1 and M2 is 1.

Question11. In a chamber, a uniform magnetic field of \( 6.5 \, \text{G} \) (where \( 1 \, \text{G} = 10^{-4} \, \text{T} \)) is maintained. An electron is shot into the field with a speed of \( 4.8 \times 10^6 \, \text{m/s} \) normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. Given: \begin{align*} e & = 1.5 \times 10^{-19} \, \text{C} \\ m_e & = 9.1 \times 10^{-31} \, \text{kg} \end{align*}

Sol: When an electron is moving perpendicular to a uniform magnetic field, it experiences a centripetal force due to the Lorentz force. This force provides the necessary centripetal force for circular motion. The path of the electron becomes a circle because the magnetic force acts as a centripetal force, constantly changing the direction of the velocity without changing its magnitude. The magnetic force \( F \) on the electron is given by: \[ F = e v B \sin \theta \] where: \begin{align*} e & = \text{charge of the electron} = 1.5 \times 10^{-19} \, \text{C} \\ v & = \text{velocity of the electron} = 4.8 \times 10^6 \, \text{m/s} \\ B & = \text{magnetic field strength} \\ \theta & = \text{angle between the velocity and magnetic field} = 90^\circ \end{align*} Since \( \sin 90^\circ = 1 \), the magnetic force becomes: \[ F = e v B \] This force acts as the centripetal force for circular motion, which is given by: \[ F_c = \frac{m_e v^2}{r} \] where: \begin{align*} m_e & = \text{mass of the electron} = 9.1 \times 10^{-31} \, \text{kg} \\ r & = \text{radius of the circular orbit} \end{align*} Equating the magnetic force to the centripetal force: \[ e v B = \frac{m_e v^2}{r} \] Solving for \( r \): \[ r = \frac{m_e v}{e B} \] First, convert the magnetic field strength from gauss to tesla: \[ B = 6.5 \, \text{G} = 6.5 \times 10^{-4} \, \text{T} \] Substitute the values into the formula: \[ r = \frac{9.1 \times 10^{-31} \times 4.8 \times 10^6}{1.5 \times 10^{-19} \times 6.5 \times 10^{-4}} \] \[ r = \frac{4.368 \times 10^{-24}}{9.75 \times 10^{-23}} \] \[ r \approx 0.448 \, \text{m} \] \[ r \approx 44.8 \, \text{cm} \] Thus, the radius of the circular orbit is approximately \( 44.8 \, \text{cm} \).

Question 12. In Exercise 4.11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.

Solution :Given:

Magnetic field strength, B = 6.5 G = 6.5 × 10-4T

Initial velocity of electron = 4.8 × 106 ms-1

Angle between the initial velocity of electron and magnetic field, θ = 900

We can relate the velocity of the electron to its angular frequency by the relation,

V = rω …(1)

Where,

V = velocity of electron

r = radius of path

ω = angular frequency

Question13. (a) A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60º with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.

(b) Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)

Solution :Given:

Number of turns in the coil, n = 30

Radius of coil, r = 8 cm

Current through the coil, I = 6.0 A

Strength of magnetic field = 1.0 T

Angle between the direction of field and normal to coil, θ = 60°

We can understand that the counter torque required to prevent the coil from rotating is equal to the torque being applied by the magnetic field.

Torque on the coil due to magnetic field is given by,

T = n × B × I × A × sinθ …(1)

Where,

n = number of turns

B = Strength of magnetic field

I = Current through the coil

A = Area of cross-section of coil

A = πr2 = 3.14 × (0.08 × 0.08) = 0.0201m2 …(2)

θ = Angle between normal to cross-section of coil and magnetic field

Now, by putting the values in equation (1) we get,

⇒ T = 30 × 6.0T × 1A × 0.0201m 2 × sin 60°

T = 3.133 Nm

Hence, the counter torque required to prevent the coil from rotating is 3.133 Nm.

b) From equation (1) we can understand that torques depend on the total area of cross-section and have no relation with the geometry of cross-section. Hence, the answer will remain unaltered if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area.

Additional Exercise of Chapter 4

Question 14. Two concentric circular coils \( X \) and \( Y \) of radii \( 16 \, \text{cm} \) and \( 10 \, \text{cm} \), respectively, lie in the same vertical plane containing the north-to-south direction. Coil \( X \) has \( 20 \) turns and carries a current of \( 16 \, \text{A} \); coil \( Y \) has \( 25 \) turns and carries a current of \( 18 \, \text{A} \). The sense of the current in \( X \) is anticlockwise, and clockwise in \( Y \), for an observer looking at the coils facing west. Give the magnitude and direction of the net magnetic field due to the coils at their center.

Sol: Given: \begin{align*} \text{Radius of coil } X & = 16 \, \text{cm} = 0.16 \, \text{m} \\ \text{Radius of coil } Y & = 10 \, \text{cm} = 0.10 \, \text{m} \\ \text{Number of turns on coil } X & = 20 \\ \text{Number of turns on coil } Y & = 25 \\ \text{Current in coil } X & = 16 \, \text{A} \\ \text{Current in coil } Y & = 18 \, \text{A} \end{align*} The magnetic field \( B_1 \) due to coil \( X \) at its center is given by: \[ B_1 = \frac{\mu_0 N_1 I_1}{2 r_1} \] where: \begin{align*} \mu_0 & = \text{permeability of free space} = 4\pi \times 10^{-7} \, \text{T m/A} \\ N_1 & = \text{number of turns on coil } X = 20 \\ I_1 & = \text{current in coil } X = 16 \, \text{A} \\ r_1 & = \text{radius of coil } X = 0.16 \, \text{m} \end{align*} Substitute the values into the formula: \[ B_1 = \frac{4\pi \times 10^{-7} \times 20 \times 16}{2 \times 0.16} \] \[ B_1 = \frac{4\pi \times 10^{-7} \times 320}{0.32} \] \[ B_1 = 4\pi \times 10^{-4} \, \text{T} \] The magnetic field \( B_2 \) due to coil \( Y \) at its center is given by: \[ B_2 = \frac{\mu_0 N_2 I_2}{2 r_2} \] where: \begin{align*} N_2 & = \text{number of turns on coil } Y = 25 \\ I_2 & = \text{current in coil } Y = 18 \, \text{A} \\ r_2 & = \text{radius of coil } Y = 0.10 \, \text{m} \end{align*} Substitute the values into the formula: \[ B_2 = \frac{4\pi \times 10^{-7} \times 25 \times 18}{2 \times 0.10} \] \[ B_2 = \frac{4\pi \times 10^{-7} \times 450}{0.20} \] \[ B_2 = 9\pi \times 10^{-4} \, \text{T} \] The net magnetic field \( B \) at the center is: \[ B = B_2 – B_1 \] \[ B = 9\pi \times 10^{-4} – 4\pi \times 10^{-4} \] \[ B = 5\pi \times 10^{-4} \approx 1.57 \times 10^{-4} \, \text{T} \] The direction of the magnetic field due to coil \( X \) is towards the observer (west direction), and due to coil \( Y \) is away from the observer (east direction). Since the currents are in opposite directions, the field due to \( Y \) subtracts from that due to \( X \). Thus, the net magnetic field due to the coils at their center is \( 1.57 \times 10^{-4} \, \text{T} \) towards the west direction.

Question15. A magnetic field of 100 G (1 G = 10−4 T) is required which is uniform in a region of linear dimension about 10 cm and area of cross-section about 10−3 m2. The maximum current-carrying capacity of a given coil of wire is 15 A and the number of turns per unit length that can be wound round a core is at most 1000 turns m−1. Suggest some appropriate design particulars of a solenoid for the required purpose. Assume the core is not ferromagnetic.

Solution :

Radius of the coil X, r1 = 16 cm = 0.16 m

Number of turns in coil X, n1 = 20

Current in the coil X, I1 = 16 A

Radius of the coil Y, r2 = 10 cm = 0.1 m

Number of turns in coil Y, n2 = 25

Current in the coil Y, I2 = 18 A

The magnetic field due to the coil X at the center is given as

Therefore, the net magnetic field is given as

B = B2 – B1 = 9π x 10-4 T – 4π x 10-4 T

= 5π x 10-4 T

= 5 x 3.14 x 10-4

= 1.57 x 10-3 T (Towards west)

Q 4.15) A magnetic field of 100 G (1 G = 10–4 T) is required, which is uniform in a region of linear dimension of about 10 cm and an area of cross-section of about 10–3 m2. The maximum current-carrying capacity of a given coil of wire is 15 A, and the number of turns per unit length that can be wound around a core is, at most, 1000 turns m–1. Suggest some appropriate design particulars of a solenoid for the required purpose. Assume the core is not ferromagnetic.

Answer:

Magnetic field strength, B = 100 G = 100 x 10-4 T

Number of turns per unit length, N = 1000 turns/m

Current carrying capacity of the coil = 15 A

Permeability of free space, μ0 = 4π x 10-7 TmA-1

The magnetic field is given as

B = μ0 NI/l

⇒ NI/l = B/μ0

= (100 x 10-4)/(4π x 10-7)

NI/l = 7961

Now, we can consider a possible combination. Let the current, I = 10 A and the length of the solenoid, l = 0.5 m

So we get

(N x 10)/0.5 = 7961

N = 398 turns ≈ 400 turns

Length about 50 cm, number of turns about 400, current about 10 A. These particulars are not unique. Some adjustment with limits is possible.

Q 4.16) For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its center is given by,

(a) Show that this reduces to the familiar result for the field at the center of the coil.

(b) Consider two parallel co-axial circular coils of equal radius R and number of turns N, carrying equal currents in the same direction, and separated by a distance R. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to R, and is given by,

Answer:

(a) Given,

(b) Let the small distance between the points P and O be d.

For the coil C, the distance O1P = x1 = (R/2) +d

For the coil d, the distance O2P = x2 = (R/2) – d

The magnetic field on the axis at a distance x from the center of the circular coil having a radius a, with n number of turns and having a current I is given by

Q 4.17 ) A toroid has a core (non-ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which 3500 turns of a wire are wound. If the current in the wire is 11 A, what is the magnetic field?
(a) outside the toroid, (b) inside the core of the toroid, and (c) in the empty space surrounded by the toroid.

Answer:

The inner radius of the core, r1= 25 cm = 0.25 m

The outer radius of the core, r2= 26 cm = 0.26 m

Number of turns of the wire, N= 3500 turns

Current in the wire, I = 11 A

(a) The magnetic field outside the toroid is zero.

(b) Inside the core of the toroid, the magnetic field induction is

B = μ0NI/l

Mean length of the toroid,

Q 4.18) Answer the following questions:
(a) A magnetic field that varies in magnitude from point to point but has a constant direction (east to west) is set up in a chamber. A charged particle enters the chamber and travels undeflected along a straight path with constant speed. What can you say about the initial velocity of the particle?
(b) A charged particle enters an environment of a strong and non-uniform magnetic field varying from point to point both in magnitude and direction and comes out of it following a complicated trajectory. Would its final speed equal the initial speed if it suffered no collisions with the environment?
(c) An electron travelling west to east enters a chamber having a uniform electrostatic field in the north-to-south direction. Specify the direction in which a uniform magnetic field should be set
up to prevent the electron from deflecting from its straight-line path.

Answer:

(a) Initial velocity is either parallel or antiparallel to the magnetic field. There is no magnetic force acting on the particle when it is parallel or antiparallel, and it moves undeflected.

(b) Yes, because magnetic force can change the direction of velocity but not its magnitude.

(c) Magnetic field should be in a vertically downward direction.

Q 4.19) An electron emitted by a heated cathode and accelerated through a potential difference of 2.0 kV enters a region with a uniform magnetic field of 0.15 T. Determine the trajectory of the electron if the field
(a) is transverse to its initial velocity, (b) makes an angle of 30º with the initial velocity.

Answer:

Magnetic field, B = 0.15 T

Potential difference, V = 2.0 kV

An electron gains kinetic energy, which is given by

E = (1/2) mv2

eV = (1/2) mv2

(a) When the field applied is transverse to the initial velocity, the force F1 acting on the electron is perpendicular to the direction of the magnetic field and the direction of motion of the electron. The perpendicular force provides the centripetal force F2 to the electron and makes it move in a circular path.

The forces F1 and F2 are equal

F1 = evB

F2 = mv2/2

F1 = F2

evB = mv2/2

⇒ r = mv/eB

The electron will move in a circular path with a radius of 1 mm in the direction perpendicular to the magnetic field.

(b) The applied force is inclined to the initial velocity. The velocity of the electron is resolved into two components, V cos θ and V sin θ. V cos θ is along the direction of the field, and it causes the electron to move in a straight line. V sin θ is along the normal, and it causes the electron to move in a circular path. Therefore, the electron moves in a helical path. The radius of the helical path is given as,

r = mV sin θ/Be

r = 0.5 mm

The helical trajectory of radius 0.5 mm along the direction of the magnetic field.

Q 4.20) A magnetic field set up using Helmholtz coils (described in Exercise 4.16) is uniform in a small region and has a magnitude of 0.75 T. In the same region, a uniform electrostatic field is maintained in a direction normal to the common axis of the coils. A narrow beam of (single species) charged particles, all accelerated through 15 kV, enters this region in a direction perpendicular to both the axis of the coils and the electrostatic field. If the beam remains undeflected when the electrostatic field is 9.0 × 10–5 V m–1, make a simple guess as to what the beam contains. Why is the answer not unique?

Answer:

Magnetic field, B = 0.75 T

Accelerating voltage, V = 15 kV = 15 x 103 V

Electrostatic field, E = 9.0 × 10–5 V m–1

Kinetic energy of the electron, E = (1/2) mv2

eV = (1/2) mv2

Therefore, (e/m) = (v2/2V)

Here,

Mass of the electron = m

Charge of the electron = e

The velocity of the electron = v

The particles are not deflected by the electric field and the magnetic field. So the force on the particle due to the electric field is balanced by force due to the magnetic field.

eE = evB

⇒ v = E/B

Therefore, (1/2) m(E/B)2 = eV

e/m = E2/2VB2

The answer is not unique because only the ratio of charge to mass is determined. Other possible answers are He++, Li++, etc.

Q 4.21) A straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended by two vertical wires at its ends. A current of 5.0 A is set up in the rod through the wires.
(a) What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero?
(b) What will be the total tension in the wires if the direction of the current is reversed, keeping the magnetic field the same as before?
(Ignore the mass of the wires.) g = 9.8 m s–2.

Answer:

Length of the rod, l = 0.45 m

Mass suspended, m = 60 g = 60 x 10-3 Kg

Current, I = 5 A

(a) Tension in the wire is zero if the force on the current-carrying wire due to the magnetic field is equal and opposite to the weight on the wire.

BIl = mg

B = mg/Il

The magnetic field set up normal to the conductor is 0. 26133 T

(b) When the direction of the current is reversed, BIl and mg will act vertically downwards; the effective tension in the wire is

T = BIl+mg

= (0. 26133 x 5.0 x 0.45) + (60 x 10-3 x 9.8) = 0.587 + (0.588) = 1.176 N

Total tension in the wire = 1.176 N

Q 4.22) The wires which connect the battery of an automobile to its starting motor carry a current of 300 A (for a short time). What is the force per unit length between the wires if they are 70 cm long and 1.5 cm apart? Is the force attractive or repulsive?

Answer:

Current in the wires, I = 300 A

Distance between the wires, d = 1.5 cm = 0.015 m

Length of the wires, l = 70 cm = 0.7 m

The force between the two wires, F = μ0I2/2πd

Here,

μ0 = permeability of the free space = 4π x 10-7 Tm/A

Q 4.23) A uniform magnetic field of 1.5 T exists in a cylindrical region of a radius of 10.0 cm, its direction parallel to the axis along east to west. A wire carrying a current of 7.0 A in the north-to-south direction passes through this region. What is the magnitude and direction of the force on the wire if,
(a) the wire intersects the axis,
(b) the wire is turned from N-S to the northeast-northwest direction
(c) the wire in the N-S direction is lowered from the axis by a distance of 6.0 cm?

Answer:

(a) Magnetic field, B = 1.5 T

Current in the wire, I = 7.0 A

Radius, r = 10 cm = 0.1 m

Diameter, l = 2 x r = 0.2 m

A force 2.1 N acts vertically downward on the wire

(a) The wire intersects the axis, θ = 900

F = BIlsinθ = 1.5 x 7 x 0.20 x sin 900

= 2.1 N

(b) The wire is turned from N-S to the northeast-northwest direction

F = BIl1Sin θ

The length of the wire after turning to the northeast-northwest direction is l1 = l/sinθ

Therefore, l = l1 sin θ

F = BIl

= 1.5 x 7 x 0.20= 2.1 N

A force 2.1 N acts vertically downward on the wire

(c) When the wire is lowered by 6 cm, then

Then,

Q 4.24) A uniform magnetic field of \( 3000 \, \text{G} \) (where \( 1 \, \text{G} = 10^{-4} \, \text{T} \)) is established along the positive \( Z \)-direction. A rectangular loop of sides \( 10 \, \text{cm} \) and \( 5 \, \text{cm} \) carries a current of \( 12 \, \text{A} \). Determine the torque and force on the loop and discuss whether the loop is in stable equilibrium. \section*

Solution: \subsection*{1. Conversion of Magnetic Field} First, convert the magnetic field from Gauss to Tesla: \[ B = 3000 \, \text{G} = 3000 \times 10^{-4} \, \text{T} = 0.3 \, \text{T} \] \subsection*{2. Magnetic Moment Calculation} The area \( A \) of the rectangular loop is: \[ A = a \times b = (10 \times 10^{-2} \, \text{m}) \times (5 \times 10^{-2} \, \text{m}) = 50 \times 10^{-4} \, \text{m}^2 \] The magnetic moment \( \vec{m} \) of the loop is given by: \[ \vec{m} = N I A \] where \( N = 1 \) (one loop), \( I = 12 \, \text{A} \), and \( A = 50 \times 10^{-4} \, \text{m}^2 \): \[ \vec{m} = 1 \times 12 \times 50 \times 10^{-4} \, \text{N m/T} = 6 \times 10^{-2} \, \text{N m/T} \] \subsection*{3. Torque Calculation} The magnetic field \( \vec{B} \) is along the \( Z \)-direction: \[ \vec{B} = 0.3 \, \text{T} \hat{k} \] The torque \( \vec{\tau} \) on the current loop is: \[ \vec{\tau} = \vec{m} \times \vec{B} \] Given \( \vec{m} \) can be along \( \hat{i} \) or \( \hat{j} \), let’s compute for each direction. \subsubsection*{(i) If \( \vec{m} \) is along \( \hat{i} \)} \[ \vec{m} = 6 \times 10^{-2} \hat{i} \] \[ \vec{\tau} = (6 \times 10^{-2} \hat{i}) \times (0.3 \hat{k}) = -1.8 \times 10^{-2} \hat{j} \, \text{N m} \] \subsubsection*{(ii) If \( \vec{m} \) is along \( \hat{j} \)} \[ \vec{m} = 6 \times 10^{-2} \hat{j} \] \[ \vec{\tau} = (6 \times 10^{-2} \hat{j}) \times (0.3 \hat{k}) = 1.8 \times 10^{-2} \hat{i} \, \text{N m} \] \subsection*{4. Force on the Loop} The net force on a current-carrying loop in a uniform magnetic field is zero because the forces on opposite sides of the loop cancel each other out: \[ \vec{F} = i (\vec{l} \times \vec{B}) = 0 \, \text{N} \] \subsection*{5. Equilibrium Analysis} – **Torque Non-Zero**: The net torque is non-zero, so the loop is not in equilibrium. – **Stable vs. Unstable**: – **Stable Equilibrium**: Occurs when the loop aligns with the magnetic field such that the potential energy is minimized. – **Unstable Equilibrium**: Occurs when the loop aligns perpendicular to the magnetic field. The torque will be zero only when the loop is aligned with the magnetic field direction. In this case, since the loop has a non-zero torque, it is in an unstable equilibrium. \section*{Final Answer} 1. **Torque**: \(1.8 \times 10^{-2} \, \text{N m}\) along \( -\hat{j} \) or \( \hat{i} \) depending on the direction of \( \vec{m} \). 2. **Force**: Zero. 3. **Equilibrium**: The loop is in an unstable equilibrium.

Q 4.25) A circular coil of 20 turns and a radius of 10 cm is placed in a uniform magnetic field of 0.10 T normal to the plane of the coil. If the current in the coil is 5.0 A, what is
(a) the total torque on the coil
(b) the total force on the coil
(c) the average force on each electron in the coil due to the magnetic
field? (The coil is made of copper wire of a cross-sectional area of 10–5 m2, and
the free electron density in copper is about 1029 m–3.)

Answer:

Number of turns, n = 20 turns

Radius of the coil, r = 10 cm = 0.1 m

Current in the coil, I = 5 A

Magnetic field strength, B = 0.10 T

The cross-sectional area of the wire, a = 10-5 m2

The angle between

Therefore, the torque is zero.

(b) Equal and opposite forces act on opposite sides of the coil. Therefore, the total force on the coil is zero.

(c) Average force on the moving electron

F = B e vd

vd is the drift velocity of the electrons

vd = I/N e A

Therefore, F = B e I/ N e a

F = B I/n a

=

Q. 4.26) A solenoid 60 cm long and of radius 4.0 cm has 3 layers of windings of 300 turns each. A 2.0 cm long wire of mass 2.5 g lies inside the solenoid (near its center) normal to its axis; both the wire and the axis of the solenoid are in the horizontal plane. The wire is connected through two leads parallel to the axis of the solenoid to an external battery which supplies a current of 6.0 A in the wire. What value of current (with an appropriate sense of circulation) in the windings of the solenoid can support the weight of the wire? g = 9.8 m s–2?

Answer:

Length of the solenoid, l = 60 cm

Layers of windings = 3

Each layer has 300 turns

Number of turns per unit length, n = (3 x 300)/0.6 = 1500 m-1

The magnetic field inside the solenoid is given by, B = μ0nI

μ0 = 4π x 10-7 T

B = (4π x 10-7) x 1500 x I

= 6π x 10-4 I———-(1)

Force due to magnetic field, F = I’Bl

I’ = Current in the wire = 6 A

l = length of the wire = 2 cm

The windings of the solenoid would support the weight of the wire when the force due to the magnetic field inside the solenoid balances the weight of the wire.

I’Bl = mg

m = mass of the wire = 2.5 g

B = mg/I’l

From equation (1), we get

6π x 10-4 I= mg/I’l

⇒ I = mg/(6π x 10-4 ) I’l

Q 4.27) A galvanometer coil has a resistance of 12 Ω, and the meter shows full-scale deflection for a current of 3 mA. How will you convert the meter into a voltmeter of range 0 to 18 V?

Solution:

Resistance of the galvanometer coil, G = 12 Ω

Current for which there is full-scale deflection, I = 3mA

A resistor with a resistance R is connected in series with the galvanometer to convert it into a voltmeter. The resistance R is given as

R = (V/Ig) – G

=

Q 4.28) A galvanometer coil has a resistance of 15 Ω, and the metre shows full-scale deflection for a current of 4 mA. How will you convert the metre into an ammeter of range 0 to 6 A?

Solution:

Resistance of the galvanometer, G = 15 Ω

Current through the galvanometer, Ig = 4 mA = 4 x 10-3 A

Ammeter range = 0 to 6 A

S = 10 mA

A 10 mA resistor must be connected in parallel to the galvanometer.

These were the solutions of Chapter 4: Moving Charges and Magnetism for the updated academic year 2024-25 for Physics students for their upcoming board examinations. Prepare the chapter well to boost your knowledge of the concepts presented in the chapter.