Question

A long solenoid having $200$ turns per cm carries a current of $1.5amp.$ At the centre of it is placed a coil of $100$ turns of cross-sectional are $3.14\times {10}^{-4}{m}^2$ having its axis parallel to the field produced by the solenoid. When the direction of current in the solenoid is reversed within $0.05sec,$ the induced e.m.f. in the coil is

• A. $0.48V$
• B. $0.048V$
• C. $0.0048V$
• D. $48V$

Answer 1

The correct option is B $0.048V$

$B={\mu }_{0}ni=(4\pi \times {10}^{-7})(200\times {10}^{-2})\times 1.5$

$=3.8\times {10}^{-2}Wb/m^2$

Magnetic flux through each turn of the coil

$\varphi =BA=(3.8\times {10}^{-2})(3.14\times {10}^{-4})=1.2\times {10}^{-5}weber$

When the current in the solenoid is reversed, the change in magnetic flux

$=2\times (1.2\times {10}^{-5})=2.4\times {10}^{-5}weber$

Induced e.m.f. $=N\frac{d\varphi }{dt}=100\times \frac{2.4\times {10}^{-5}}{0.05}=0.048V$.