A coil 3.65 cm radius, containing 550 turns, is placed in a uniform magnetic fie

Question

A coil 3.65 cm radius, containing 550 turns, is placed in a uniform magnetic field that varies with time according to B = (1.20 Times 10^-2 T/s)t + (2.75 Time 10^-5 T/s^4)t^4. The coil is connected to a 540 Ohm resistor, and its plane is perpendicular to the magnetic field. You can ignore the resistance of the coil. Find the magnitude of the induced emf in the coil as a function of time. Epsilon = 8.79 Times 10^-3 V +(8.06 Times 10^-5 V/s^3)t^3 epsilon = 2.76 Times 10^-2 V +(6.33 Times 10^-5 V/s^3)t^3 epsilon = 2.76 Times 10^-2 V +(2.53 Times 10^-4 V/s^3)t^3 epsilon = 8.79 Times 10^-3 V +(2.53 Times 10^-4 V/s^3)t^3 What is the current in the resistor at time t_0 = 5.35 s ?

Explanation / Answer

a) The magnetic flux through one loop of the coil is B(t) = pi*r^2*B(t),
so the magnitude of the induced emf in the coil is

E(t) = N*pi*r^2 (Ba + 4Bbt^3)

= 550*pi*(0.0365m)^2*[(0.0120T/s) + 4(2.75*10^-5T/s4)t^3]
= 2.3019[(0.0120T/s) + 4(2.75*10^-5T/s4)t^3]
= 2.76*10^-2 V + (2.53*10^-4 V/s^3)t^3

b)
At time t0 = 5.35s, the magnitude of the induced emf is

E(t0) = 2.76*10^-2 V + (2.53*10^-4 V/s^3)(5.35s)^3
= 0.06634 V

So the current through the resistor at this time is

i(t0) = E(t0)/R
= 0.06634/540
= 0.0001228 A

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