Home
Class 12
PHYSICS
A circular coil of radius 4 cm having 50...

A circular coil of radius 4 cm having 50 turns carries a current of 2 A . It is placed in a uniform magnetic field of the intensity of `0.1` Weber `m^(-2)` . The work done to rotate the coil from the position by `180^@` from the equilibrium position

A

0.1 J

B

0.2 J

C

0.4 J

D

0.8 J

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to calculate the work done in rotating a circular coil in a magnetic field. Here are the steps to arrive at the solution: ### Step 1: Calculate the Area of the Coil The area \( A \) of a circular coil is given by the formula: \[ A = \pi r^2 \] where \( r \) is the radius of the coil. Given that the radius \( r = 4 \) cm, we first convert it to meters: \[ r = 4 \, \text{cm} = 0.04 \, \text{m} \] Now, substituting the value of \( r \): \[ A = \pi (0.04)^2 = \pi (0.0016) = 0.0016\pi \, \text{m}^2 \] ### Step 2: Calculate the Magnetic Moment The magnetic moment \( \mu \) of the coil is given by: \[ \mu = n \cdot I \cdot A \] where \( n \) is the number of turns, \( I \) is the current, and \( A \) is the area calculated in Step 1. Given \( n = 50 \) turns and \( I = 2 \) A: \[ \mu = 50 \cdot 2 \cdot (0.0016\pi) = 100 \cdot 0.0016\pi = 0.16\pi \, \text{A m}^2 \] ### Step 3: Calculate the Work Done The work done \( W \) in rotating the coil in a magnetic field is given by the change in potential energy: \[ W = \mu B (\cos \theta_i - \cos \theta_f) \] where: - \( B \) is the magnetic field intensity (given as \( 0.1 \, \text{Wb/m}^2 \)), - \( \theta_i \) is the initial angle (0 degrees), - \( \theta_f \) is the final angle (180 degrees). Substituting the values: \[ \cos \theta_i = \cos(0) = 1 \] \[ \cos \theta_f = \cos(180) = -1 \] Thus, \[ W = \mu B (1 - (-1)) = \mu B (1 + 1) = 2\mu B \] ### Step 4: Substitute the Values Now substituting \( \mu = 0.16\pi \) and \( B = 0.1 \): \[ W = 2 \cdot (0.16\pi) \cdot (0.1) = 0.032\pi \, \text{J} \] ### Step 5: Calculate the Numerical Value Using \( \pi \approx 3.14 \): \[ W \approx 0.032 \cdot 3.14 \approx 0.10048 \, \text{J} \] ### Final Answer The work done to rotate the coil by \( 180^\circ \) from the equilibrium position is approximately: \[ W \approx 0.1 \, \text{J} \]
Promotional Banner

Similar Questions

Explore conceptually related problems

A circular coil of radius 10 cm having 100 turns carries a current of 3.2 A. The magnetic field at the center of the coil is

A circular coil of radius 10 cm "and" 100 turns carries a current 1A. What is the magnetic moment of the coil?

A circular coil of radius 1*5 cm carries a current of 1*5A . If the coil has 25 turns , find the magnetic field at the centre.

A circular coil of radius 4 cm has 50 turns. In this coil a current of 2 A is flowing. It is placed in a magnetic field of 0.1 weber//m^(2). The amount of work done in rotating it through 180^(@) from its equilibrium position will be

A coil having N turns carry a current I as shown in the figure. The magnetic field intensity at point P is

A rectangular loop of area 5m^(2) , has 50 turns and carries a current of 1A. It is held in a uniform magnetic field of 0.1T, at an angle of 30^(@) . Calculate the torque experienced by the coil.

A circular coil having 50 turns, each with an area of 0.01 m, carries a current of 2 A. Calculate its magnetic dipole moment.

A 100 turn closely wound circular coil of radius 10cm carries a current of 3*2A .What is the field at the centre of the coil?

A tightly wound 100 turn coil of radius 10cm is carrying a current of 1A . What is the magnitude of the magnetic field at the centre of the coil?

A rectangular coil 20cmxx20cm has 100 turns and carries a current of 1 A . It is placed in a uniform magnetic field B=0.5 T with the direction of magnetic field parallel to the plane of the coil. The magnitude of the torque required to hold this coil in this position is