A rectangular loop has a sliding connector P Q of length l and resistance R Ω and it is moving with a speed v as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents I 1 , I 2 a n d I are

Question

A rectangular loop has a sliding connector P Q of length l and resistance R Ω and it is moving with a speed v as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents I 1 , I 2 a n d I are, I 1 = - I 2 = B l v R , I = 2 B l v R, I 1 = I 2 = B l v 3 R , I = 2 B l v 3 R, I 1 = I 2 = I = B l v R, I 1 = I 2 = B l v 6 R , I = B l v 3 R

MARKS App · Accepted Answer

A moving conductor is equivalent to battery of emf = v B l (motion emf) Equivalent circuit I = I 2 + I 2 Applying Kirchhoff's law I 1 R + I R - v B l = 0 … ( i ) I 2 R + I R - v B l = 0 ...(ii) Adding Eqs. (i) and (ii), we get 2 I R + I R = 2 v B l I = 2 v B l 3 R I 1 = I 2 = v B l 3 R

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