Electromagnetic Induction 63. Let the rod has a velocity v at any instant, Then, at the point, e = Bℓv Now, q = c × potential = ce = CBℓv Current I =
B
dq d CBlv dt dt
l v
dv (where a acceleration) CBla dt From figure, force due to magnetic field and gravity are opposite to each other. So, mg – IℓB = ma 2 2 mg – CBℓa × ℓB = ma ma + CB ℓ a = mg mg 2 2 a(m + CB ℓ ) = mg a= m CB2 64. a) Work done per unit test charge (E = electric field) = E. dl E. dl = e
mg
= CBl
d E dl = dt E 2r = r 2
dB E 2r = A dt dB dt
r 2 dB r dB 2 dt 2 dt b) When the square is considered, E dl = e E=
E × 2r × 4 =
dB (2r )2 dt
dB 4r 2 r dB E= dt 8r 2 dt The electric field at the point p has the same value as (a). E=
65.
di = 0.01 A/s dt di = 0.02 A/s dt n = 2000 turn/m, R = 6.0 cm = 0.06 m r = 1 cm = 0.01 m a) = BA d di 0nA dt dt –7 3 –4 –2 = 4 × 10 × 2 × 10 × × 1 × 10 × 2 × 10 2 –10 = 16 × 10 –10 = 157.91 ×10 –8 = 1.6 × 10 d or, for 1 s = 0.785 . dt
For 2s
b) E.dl =
–4
[A = × 1 × 10 ]
d dt 38.17
P r