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Standard XII

Physics

Motional EMF

There is a un...

Question

There is a uniform magnetic field B normal to the xy plane. A conductor ABC has length $A B = l_{1}$ , parallel to the x-axis, the length $B C = l_{2}$ , parallel to the y axis. ABC moves in the xy plane with velocity $v_{x}^i + v_{y}^j$ . The potential difference between A and C is proportional to :

v_{x} l_{1} + v_{y} l_{2}

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v_{x} l_{2} + v_{y} l_{1}

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v_{x} l_{2} - v_{y} l_{1}

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v_{x} l_{1} - v_{y} l_{2}

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Solution

The correct option is C $v_{x} l_{2} - v_{y} l_{1}$
Let magnetic field be $B^k$
$\to v = v_{x}^i + v_{y}^j, V_{B} - V_{A} = (l_{1}^i) . (\to v \times B^k) = B v_{y} l_{1} V_{C} - V_{B} = (l_{2}^j) . (\to v \times B^k) = - B v_{x} l_{2} \Rightarrow V_{A} - V_{C} = B v_{x} l_{2} - B v_{y} l_{1}$

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There is a uniform magnetic field B normal to the xy plane. A conductor ABC has length AB=l1, parallel to the x-axis, the length BC=l2, parallel to the y axis. ABC moves in the xy plane with velocity vx^i+vy^j. The potential difference between A and C is proportional to :

The correct option is C vxl2−vyl1Let magnetic field be B^k →v=vx^i+vy^j,VB−VA=(l1^i).(→v×B^k)=Bvyl1VC−VB=(l2^j).(→v×B^k)=−Bvxl2⇒VA−VC=Bvxl2−Bvyl1

The correct option is C $v_{x} l_{2} - v_{y} l_{1}$
Let magnetic field be $B^k$
$\to v = v_{x}^i + v_{y}^j, V_{B} - V_{A} = (l_{1}^i) . (\to v \times B^k) = B v_{y} l_{1} V_{C} - V_{B} = (l_{2}^j) . (\to v \times B^k) = - B v_{x} l_{2} \Rightarrow V_{A} - V_{C} = B v_{x} l_{2} - B v_{y} l_{1}$