A metal wire PQ slides on parallel metallic r | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The induced electromotive force $(e.m.f.)$ in the moving wire is given by $\varepsilon = B \ell v$,where $B = 0.5 \ T$,$\ell = 0.25 \ m$,and $v$ i

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(B) The induced electromotive force $(e.m.f.)$ in the moving wire is given by $\varepsilon = B \ell v$,where $B = 0.5 \ T$,$\ell = 0.25 \ m$,and $v$ is the speed.So,$\varepsilon = 0.5 	imes 0.25 	imes v = 0.125v \ V$.The net current $I$ in the circuit is $I = \frac{E - \varepsilon}{R} = \frac{10 - 0.125v}{2} \ A$.The magnetic force acting on the wire is $F_m = B I \ell = 0.5 	imes I 	imes 0.25 = 0.125I \ N$.Since the wire moves at a constant speed,the external force must balance the magnetic force: $F_{ext} = F_m = 0.5 \ N$.Substituting $I$ into the force equation: $0.5 = 0.125 	imes \left( \frac{10 - 0.125v}{2} ight)$.$0.5 	imes 2 = 0.125 	imes (10 - 0.125v) \implies 1 = 1.25 - 0.015625v$.$0.015625v = 0.25 \implies v = \frac{0.25}{0.015625} = 16 \ m/s$.

Similar Questions

Prove that the mechanical power required to move a rod in a uniform magnetic field is converted into electrical power.

$A$ rod of length $l$ is rotated with angular velocity $\omega$ about one of its ends,in a region perpendicular to a uniform magnetic field of induction $B$. The induced e.m.f. in the rod is:

$A$ conductor of $3 \ m$ in length is moving perpendicularly to a magnetic field of $10^{-3} \ T$ with a speed of $10^2 \ m/s$. The $e.m.f.$ produced across the ends of the conductor will be ........ $V$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module