A wire frame PQRSTU is moving horizontally wi | vedclass

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(A) The induced emf in a conductor of length $l$ moving with velocity $v$ in a magnetic field $B$ is given by $\varepsilon = Bvl \sin 	heta$, where $

Vedclass · Accepted Answer

the magnitude of induced emf between $P$ and $Q$ is $Bv\left( \frac{2L}{3} \right)$

Vedclass · Answer

(A) The induced emf in a conductor of length $l$ moving with velocity $v$ in a magnetic field $B$ is given by $\varepsilon = Bvl \sin 	heta$, where $l$ is the length of the conductor perpendicular to the velocity vector.For the segment $PQ$, the vertical length is $L/3$. Thus, the induced emf is $\varepsilon_{PQ} = Bv(L/3)$. Therefore, option $B$ is correct and option $A$ is incorrect.In a moving conductor, the motional electric field is $E = v 	imes B$. For the vertical segment $RS$, the length is $L$, so there is a potential difference across it, meaning the electric field is non-zero. Thus, option $C$ is correct.For the segment $QP$, although it is a horizontal wire, it is moving in a magnetic field. The motional electric field $E = v 	imes B$ exists within the material of the wire due to the Lorentz force on the free electrons. Thus, the electric field in the portion $QP$ is also non-zero. Thus, option $D$ is correct.Since the question asks for the $INCORRECT$ statement, the answer is $A$.

$A$ wire frame $PQRSTU$ is moving horizontally with velocity $v$ in a uniform magnetic field $B$ acting perpendicular to its plane as shown in the figure. Choose the $INCORRECT$ statement.

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