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Electromagnetic Induction Physics Ncert Solutions Class 12th Physics Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Six

A magnetic field B = B₀sin (ωt)k covers a large region where a wire AB slides smoothly over two parallel conductors separated by a distance d (figure). The wires are in the x-y plane. The wire AB (of length d) has resistance R and the parallel wires have negligible resistance. If AB is moving with velocity v, what is the current in the circuit. What is the force needed to keep the wire moving at constant velocity?

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Payal Gupta | Contributor-Level 10

4 months ago

This is a long answer type question as classified in NCERT Exemplar

Let us assume that parallel wires at are y=0 i.e along x-axis and y=d.at t=0, AB has x=0 i.e along y-axis and moves with a velocity v

Motional emf across AB is = (B₀sinwt)vd (-j)

Emf due to change in field along OBAC = -B₀wcoswt x (t)d

Total emf in the circuit = emf due to change in field (along OBAC)+ the motional emf across AB= - B₀d {wx coswt+v sinwt}

Electric current in clockwise direction is given by = $\frac{B_{0} d}{R}$ (wxcoswt+vsinwt)

The force acting on the conductor is given by F= iLbsin90=iLb

Substituting the values

F= $\frac{B_{0} d}{R}$ (wx coswt+v sinwt) × d &time

...more

This is a long answer type question as classified in NCERT Exemplar

Let us assume that parallel wires at are y=0 i.e along x-axis and y=d.at t=0, AB has x=0 i.e along y-axis and moves with a velocity v

Motional emf across AB is = (B₀sinwt)vd (-j)

Emf due to change in field along OBAC = -B₀wcoswt x (t)d

Total emf in the circuit = emf due to change in field (along OBAC)+ the motional emf across AB= - B₀d {wx coswt+v sinwt}

Electric current in clockwise direction is given by = $\frac{B_{0} d}{R}$ (wxcoswt+vsinwt)

The force acting on the conductor is given by F= iLbsin90=iLb

Substituting the values

F= $\frac{B_{0} d}{R}$ (wx coswt+v sinwt) × d × B₀sinwt

= $\frac{B_{0}^{2} d^{2}}{R}$ (wx coswt+v sinwt)sinwt

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