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

Find the current in the sliding cod AB (resistance = R) for the arrangement shown in figure. B is constant and is out of the paper. Parallel wires have no resistance, v is constant. Switch S is closed at time t = 0.

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Payal Gupta | Contributor-Level 10

4 months ago

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

The conductor of length d moves with speed v, perpendicular to magnetic field B as shown in figure. This produces motional emf across two ends of rod, which is given by= vBd.

Since, switch S is closed at time t= 0. current start growing in inductor by the potential

Difference due to motional emf.

Applying Kirchhoff’s voltage rule, we have

-L $\frac{d I}{d t}$ +vBd=IR

On solving differential equation we get I= $\frac{v B d}{R}$ + Ae^-Rt/2

At t=0 I=0

A= -vBd/R

I= $\frac{v B d}{R}$ (1-e^-Rt/L)

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alok kumar singh

B_v = B sin 60°

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Figure shows a circuit that contains four identical resistors with resistance R = $2.0 Ω,$ two identical inductors with inductance L = 2.0mH and an ideal battery with emf E = 9V. The current ‘i’ just after switch ‘S’ is closed will be:

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R = $2 Ω$

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Find the current in the sliding cod AB (resistance = R) for the arrangement shown in figure. B is constant and is out of the paper. Parallel wires have no resistance, v is constant. Switch S is closed at time t = 0.

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