123 E Blv = R R ic = 0 1 1 Uc = CE2 = CB2l2v2 2 2
\ fm = B tan q x 2 df dx e = - m = - 2B tan q x dt dt = 2B tan q vx R = r l = r (2x tan q) where, r = resistance per unit length of the conductor. e Bv \ i= = = constant. R r
iR =
3. Apply Fleming’s left hand rule. 4. For SHM, v = w A cos wt e = Blv = Blw A cos wt ì ï e0 cos wt e =í ï- e0 cos wt î
for nT < t < (2n - 1) for
T 2
(2n - 1)T > t > nT 2
5. fm = BA At any instant when wires have moved through a distance x, A = ( a + 2 x )2 fm = B( a + 2x )2 df dx |e| = m = 4B( a + 2x ) dt dt = 4B( a + 2x ) v0 |e| 4B ( a + 2x ) v0 Bv0 |i| = = = R l ´ 4 ( a + 2x ) l 6. A = l2 dA dl = 2l = - 2 la dt dt
at
dl ö æ ça = - ÷ dt è ø
fm = BA df dA e=- m =-B = 2Bla dt dt l=a e = 2a aB
7. At this instant, direction of motion of wire PQ is perpendicular to its length. e = Blv \ 8. q = CV = CBlv = 20 ´ 10-6 ´ 0.5 ´ 0.1 ´ 0.2 = 0.2 mC Plate A is positive while plate B is negative. æ1 ö 9. fm = BA = B ç lx ÷ è2 ø l v q
But l = 2x tan q
x
10. fm = BA cos q = BA cos wt df e = - m = w BA sin wt dt 2 But A=b e = b2Bw sin wt \ 11. Induced emf dB = (1)2 ´ 2 ´ 10-3 dt = 2 ´ 10-3 V W = qe = 1 ´ 10-6 ´ 2 ´ 10-3 = 2 ´ 10-9 J e = a2
12. In the steady capacitor = 0.
state,
current
through
20 =4A 5 f1 = 0, f2 = iL L = 4 ´ 500 ´ 10-2 = 2 Wb Df = f2 - f1 = 2 Wb. 1 1 æ1 ö 13. Li2 = ç Li02 ÷ 2 2 è2 ø i0 i= 2 t æ - ö i i0 ç1 - e t ÷ = 0 ç ÷ 2 è ø 2 -1 e -t / t = 2 æ 2 ö÷ t = t ln ç ç 2 - 1÷ è ø æ L 2 ö÷ = ln ç ç R è 2 - 1 ÷ø m i 14. B = 0 2pa m iqv F = qvB = 0 2pa iL =
15. Consider an elementary section of loop of width dx at a distance x from wire AB