Electromagnetic Induction — Easy JEE physics MCQ
A coil of $100$ turns has magnetic flux changing from $0.02$ Wb to $0.04$ Wb in $0.1$ s. The induced EMF is:
- A. $20$ V
- B. $2$ V
- C. $200$ V
- D. $0.2$ V
Solution
$\varepsilon = -N\frac{d\phi}{dt} = 100 \times \frac{0.04 - 0.02}{0.1} = 100 \times 0.2 = 20$ V
