Sign in
Please select an account to continue using cracku.in
↓ →
Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
A straight wire AB of mass $$40$$ g and length $$50$$ cm is suspended by a pair of flexible leads in uniform magnetic field of magnitude $$0.40$$ T as shown in the figure. The magnitude of the current required in the wire to remove the tension in the supporting leads is _____ A. (Take $$g = 10$$ m s$$^{-2}$$.
Correct Answer: 2
For the wire to be in equilibrium with zero tension in the supporting leads, the magnetic force must balance gravity.
$$BIL = mg$$
Given: $$m = 40$$ g $$= 0.04$$ kg, $$L = 50$$ cm $$= 0.5$$ m, $$B = 0.40$$ T, $$g = 10$$ m/s².
$$0.40 \times I \times 0.5 = 0.04 \times 10$$
$$0.2I = 0.4$$
$$I = 2 \text{ A}$$
The current required is $$\mathbf{2}$$ A.
Predict your JEE Main percentile, rank & performance in seconds
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.