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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

Answer :

oyejam

Answer:

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

1) The maximum induced EMF in the loop of wire is; EMF_max = 9.52 × 10^(-4) V

2) The maximum current through the bulb is;

I_max = 1.36 × 10^(-4) A

We are given;

Number of turns; N = 1

Magnitude of magnetic field; B = 2 T

Area; A = 2 cm² = 0.0002 m²

Angular frequency; ω = 143 /min = 2.38 /s

Resistance; R = 7 Ω.

1) Formula for maximum induced EMF is;

EMF_max = NAωB

Plugging in the relevant values gives;

EMF_max = 1 × 0.0002 × 2.38 × 2

EMF_max = 9.52 × 10^(-4) V

2) Formula for maximum current through the bulb is given as;

I_max = EMF_max/R

Plugging in the relevant values;

I_max = (9.52 × 10^(-4))/7

I_max = 1.36 × 10^(-4) A

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