Question

A coil is placed perpendicular to a magnetic field of 5000 T. When the field is changed to 3000 T in 2s, an induced emf of 22 V is produced in the coil. If the diameter of the coil is 0.02 m, then the number of turns in the coil is :

• A. 7
• B. 70
• C. 35
• D. 140

Answer 1

The Correct Option is B

The induced emf in the coil is given by:

\[ \mathcal{E} = N \left( \frac{\Delta \phi}{t} \right) \]

Where:

\[ \Delta \phi = (\Delta B)A \]

Given:

\[ B_i = 5000 \, \text{T}, \quad B_f = 3000 \, \text{T} \] \[ d = 0.02 \, \text{m} \implies r = 0.01 \, \text{m} \]

Calculating the change in magnetic flux:

\[ \Delta \phi = (\Delta B)A = (2000)(\pi)(0.01)^2 = 0.2\pi \]

Substituting values into the emf equation:

\[ \mathcal{E} = N \left( \frac{0.2\pi}{2} \right) \]

Given \( \mathcal{E} = 22 \, \text{V} \), we get:

\[ 22 = N \left( \frac{0.2\pi}{2} \right) \]

Solving for \( N \):

\[ N = 70 \]