JEE WhatsApp Group
Sign in
Please select an account to continue using cracku.in
↓ →
JEE WhatsApp Group
Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
A coil is placed perpendicular to a magnetic field of $$5000$$ T. When the field is changed to $$3000$$ T in $$2$$ s, 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:
$$|emf| = N\frac{d\Phi}{dt} = N \cdot A \cdot \frac{\Delta B}{\Delta t}$$.
$$A = \pi(0.01)^2 = \pi \times 10^{-4}$$. $$\Delta B = 5000-3000 = 2000$$ T. $$\Delta t = 2$$ s.
$$22 = N \times \pi \times 10^{-4} \times 1000 = N \times 0.1\pi$$.
$$N = \frac{22}{0.1\pi} = \frac{220}{\pi} \approx 70$$.
The answer is Option (2): 70.
Click on the Email ☝️ to Watch the Video Solution
Create a FREE account and get:
Educational materials for JEE preparation
Ask our AI anything
AI can make mistakes. Please verify important information.
AI can make mistakes. Please verify important information.
