Posted by **MPHO** on Tuesday, May 1, 2012 at 12:51am.

A 5.0 cm diameter coil of 20 turns with a resistance of 0.50 Ù is in a time-varying magnetic field which is perpendicular to the coil and is given by the expression;

B = 0.010t2 + 0.020t, where B is in tesla and t is in seconds.

a) Draw a graph of B as a function of time from t = 0 s to t = 10 s.

b) Write down an expression for the induced current I(t) as a function of time.

c) Evaluate I at t = 5 s and t = 10 s.

- PHYSICS -
**Elena**, Tuesday, May 1, 2012 at 8:21am
(a) parabolic curve,

(b)

ε = -N•(dΦ/dt)= -N•(d(B•A•cosα)/dt),

cosα =1, A =π•d^2/4.

ε = I•R,

I•R = -N• A •(dB/dt),

I =- (N• π•d^2/4•R) •(0.01•2•t +0.02) =

= - (20• π•0.25•0.02/4•0.5)(t+1) =0.157(t+1),

(c)

t=5 s => I = 0.157•6 =0.942 A,

t= 10 s => I = 0.157•11=1.727 A.

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