Question

A field of strength $\frac{5\times {10}^4}{\pi }$ ampere turns /meter acts at right angles to a coil of 50 turns of area ${10}^{-2}$ $m^2$. The coil is removed from the field in 0.1 second. Then, the induced e.m.f in the coil is:

• A. 0.1 V
• B. 0.2 V
• C. 1.96 V
• D. 0.98 V

Answer 1

The correct option is A 0.1 V

$Emf=\frac{BAn}{t}$
$=\frac{5\times {10}^4}{\pi }\times \frac{{10}^{-2}\times 50}{0.1}\times 4\pi \times {10}^{-7}$
$(\because B={\mu }_{o}H)$
where H is in Ampere/ meter
$(and{\mu }_o=4\pi \times {10}^{-7})=0.1$ V