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(B) The motional electromotive force (emf) induced in a conductor of length $l$ moving with velocity $v$ in a magnetic field $B$ is given by the formu

Vedclass · Accepted Answer

$V_A-V_B=4 	ext{ V}$

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(B) The motional electromotive force (emf) induced in a conductor of length $l$ moving with velocity $v$ in a magnetic field $B$ is given by the formula $\varepsilon = l(v 	imes B)$.Here,the length of the rod $l = 50 	ext{ cm} = 0.5 	ext{ m}$.The velocity $v = 8 	ext{ ms}^{-1}$ and the magnetic field $B = 2 	ext{ T}$.The angle between the magnetic field and the normal to the plane of motion is $	heta = 90^{\circ} - 60^{\circ} = 30^{\circ}$.The induced emf is $\varepsilon = B v l \sin(	heta)$.Substituting the values:$\varepsilon = 2 	imes 8 	imes 0.5 	imes \sin(30^{\circ})$$\varepsilon = 8 	imes 0.5 = 4 	ext{ V}$.Thus,the potential difference between the ends is $4 	ext{ V}$. Based on the direction of motion and the magnetic field,$V_A - V_B = 4 	ext{ V}$.

$A$ metal rod $AB$ of length $50 \text{ cm}$ is moving at a velocity $8 \text{ ms}^{-1}$ in a magnetic field of $2 \text{ T}$. If the field is at $60^{\circ}$ with the plane of motion as shown in the figure,then the potentials $V_A$ and $V_B$ are related by

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