A loop of wire in the shape of a rectangle rotates with a frequency of 143 rotation per minute in an applied magnetic field of magnitude 2 T. Assume the magnetic field is uniform. The area of the loop is A = 2 cm2 and the total resistance in the circuit is 7 Ω. 1. Find the maximum induced emf.
e m fmax =
2. Find the maximum current through the bulb.
Imax

1. e m fmax = 0.00598 Volt

2. Imax = 0.000854 Amp

Explanation:

1. Find the maximum induced emf.

e m fmax =

Given that e m fmax = N*A*B*w

N = 1

A = 2 cm^2 = 0.0002 m^2

f = 143 rotation per minute = 143/min

f = (143/min) * (1 min/60 sec) = 2.38/sec

w = 2Πf = 2 * Π * 2.38 = 14.95 rad/sec

B = 2T

e m fmax = N*A*B*w

e m fmax = 1 * 0.0002 * 2 * 14.95

e m fmax = 0.00598 Volt.

2. Find the maximum current through the bulb.

Imax = e m fmax / R

Where R is the total resistance in the circuit is 7 Ω.

Imax = 0.00598/7 = 0.000854 Amp.

Imax = 0.000854 Amp

10m/s

Explanation:

Using the law of conservation of momentums

M1u1+m2u2 = (m1+m2)v

Substitute.

4000(10)+1500(10) = (4000+1500)v

40,000+15,000 = 5,500v

55000 = 5500v

v = 55000/5500

v= 10m/s

Hence the velocity of the truck after Collision is 10m/s