Post a New Question


posted by .

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.


    (a) parabolic curve,
    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),
    t=5 s => I = 0.157•6 =0.942 A,
    t= 10 s => I = 0.157•11=1.727 A.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question