Question

A square wire loop with 3.00m sides is perpendicular to a uniform magnitude field, with half the area of the loop in the field, as shown below. The loop contains a 20.0 volt battery with a 210.0 ohm resistance in the circuit. If the magnitude of the field varies with time according to B(t0 = (9.00T/s)t - (3.60 T/s^2)*t^2. What is the induced emf in the circuit at 2.00 seconds? What is the total emf in the circuit at 2.00 seconds? What is the direction of the total current at that time (2 seconds)? What is the total current at that time (2 seconds)?

Answer 1

$\phi (B) = B(t) . A$

E = induced EMF = $-d\phi _{B}/dt$ = - A dB(t)/dt

= - (1/2) X 3m X 3m X [ 9.00 - 7.2 t]

= -4,5 X [ 9-7.2t]

Since dB/dt is negative , induced emf must oppose the reduction .

[Note:The diagram in this problem does not give exact direction of B field( It should normally be represented by X or . ) It is assumed B is out of the plane of the screen and since it is reducing, the induced current should be such as to produce B to increase in outward direction and hence should be anti-clockwise]

b . So total emf at t=2 is = 20v - 24.3 V = - 4.3V

c Total current will be anti-clockwise

d. Total current -4.3v/210 ohms = -0.20 Ameres