A long solenoid with 30 turns/cm has a small loop of area $3\, cm^2$ placed inside it normal to its axis. If the current carried by the solenoid changes steadily from 3.0 A to 6.0 A in 0.1 s, the induced emf in the loop is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → $108π × 10^{-7} V$
Given:
Number of turns per unit length: $n = 30 \,\text{turns/cm} = 3000 \,\text{turns/m}$
Area of loop: $A = 3 \,\text{cm}^2 = 3 \times 10^{-4} \,\text{m}^2$
Current changes: $\Delta I = 6.0 - 3.0 = 3.0 \,\text{A}$
Time: $\Delta t = 0.1 \,\text{s}$
Permeability of free space: $\mu_0 = 4\pi \times 10^{-7} \,\text{H/m}$
Magnetic field in solenoid: $B = \mu_0 n I$
Change in magnetic field: $\Delta B = \mu_0 n \Delta I$
Induced emf: $E = \frac{\Delta \Phi}{\Delta t} = \frac{A \Delta B}{\Delta t}$
$E = \frac{A (\mu_0 n \Delta I)}{\Delta t}$
$E = \frac{(3 \times 10^{-4})(4\pi \times 10^{-7})(3000)(3)}{0.1}$
$E = \frac{(3 \times 10^{-4})(36\pi \times 10^{-4})}{0.1}$
$E = \frac{108\pi \times 10^{-8}}{0.1}$
$E = 108\pi \times 10^{-7}$