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3 years ago

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A 24-cm circumference loop of wire has a resistance of 0.14 Ω. The loop is placed between the poles of an electromagnet, and a f

ield of 0.55 T is switched on in a time of 15 ms. What is the induced current in the loop?

1 answer:

Triss [41]3 years ago

4 0

Answer:

The induced current in the loop is 1.2 A

Explanation:

Given;

length of the wire, L = 24 cm = 0.24 m

resistance of the wire, R = 0.14 Ω.

magnetic field strength, B = 0.55 T

time, t = 15 ms = 15 x 10⁻³ s

Circumference of a circle is given as;

L = 2πr

0.24 = 2πr

r = 0.24 / 2π

r = 0.0382 m

Area of a loop is given as;

A = πr²

A = π (0.0382)²

A = 0.004585 m²

Induced emf is given as;

$emf = -\frac{\delta \phi}{dt}$

Ф = ΔB x A

Ф = ( 0 - 0.55 T) x 0.004585 m²

Ф = -0.002522 T.m²

$emf = - (\frac{-0.002522}{15*10^{-3}} )\\\\emf = 0.168 \ V$

According to ohm's law;

V = IR

Where;

I is current

The induced current in the loop is calculated as;

I = V / R

I = 0.168 / 0.14

I = 1.2 A

Therefore, the induced current in the loop is 1.2 A

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where

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$\omega_i = 11.0 rev/s$ is the initial angular speed

$\alpha=2.79 rev/s^2$ is the angular acceleration

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$\alpha = \frac{\omega_f-\omega_i}{t}$

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$\omega_i = 0 rev/s$ is the initial angular speed

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$\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{(11.0 rev/s)^2-0^2}{2(2.79 rev/s^2)}=21.7 rev$

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4 years ago