2. According to Lenz’s law, the current induced will oppose the change in flux through the loop.

a. Here area of the loop increase with time, the flux through it increases, the current induced will be such that its magnetic field produced is in the opposite direction.

Hence, according to right hand rule, the current will be in the clockwise direction.

b. Since the area of the loop decreases with time, the flux through it decreases and hence the current induced will be such that its magnetic field produced is in the same direction to maintain the flux right hand rule, direction of current, adcb.

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3. e = dΦ = d (BA) = A (µ0) n di = 2 x 10-4 x 4 π x 10-7 x 1500 x 2  = 7.54 x 10-6 V

          dt     dt                    dt                                             0.1

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4. (i)  e = B l v = 0.3 x 0.08 x 0.01 = 2.4 x 10-4 V and t = 0.02/ 0.01 = 2s

    (ii) e = B b v = 0.3 x 0.02 x 0.01 = 0.6 x 10-4 V and t = 0.08/0.01 = 8s

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5. e = B l 2 ω = 0.5 x 12 x 400  = 100 V

             2              2

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6. I = e = N ω AB = 20 x 50 x π x 0.08 x 0.08 x 3 x 10-2 = 0.0603 A

        R        R                             10     

    P = e l  = 0.603 x 0.0603 = 0.018 W

          2             2

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7. (a) e = B l  v = 0.3 x 10-4 x 5 x 10 = 1.5 x 10-3 V

    (b) Fleming’s right hand thumb rule gives direction of induced current. Here it is west to east.

     (c) Eastern End of wire have higher potential.

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8. e = L di/dt and so     L = e  = 200 = 4 H

                                          di     5

                                          dt    0.1

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9. e = dΦ  and e = µ d l          so   dΦ = µ d l    =  1.5 x 20 = 30 Wb

          dt                  dt                 dt    dt

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10. Bv = B sin δ = 5 x 10-4 sin300 = 2.5 x 10 – 4 T

     e = (Bv ) l v = (B sin δ) x l  x v = (5 x 10-4 sin300 ) x 25 x 500 = 3.125 V