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

Answer: (B)

70

Answer 2

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 $N$:

$N = 70$