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Found in: Page 813

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# (a) What is the maximum torque on a 150-turn square loop of wire 18.0 cm on a side that carries a 50.0-A current in a 1.60-T field? (b) What is the torque when θ is 10.9º?

(a) The maximum torque is $360\mathrm{N}‐\mathrm{m}$.

(b) When $\mathrm{\theta }=10.9°$ , the torque equals $68.1\mathrm{N}‐\mathrm{m}$.

See the step by step solution

## Step 1: Definition of Torque.

Torque can be a force applied on the different current carrying loops/segments in a uniform magnetic field.

The torque can be estimated using the expression,

${\mathbf{\tau }}{\mathbf{=}}{\mathbf{NIABsin}}{\mathbf{\left(}}{\mathbf{\theta }}{\mathbf{\right)}}{\mathbf{}}$ ………………(1)

where the number of turns, ${\mathbf{N}}$ , the current, ${\mathbf{I}}$ , the area of the loop, ${\mathbf{A}}$ , and the inclination angle of the loop, ${\mathbf{\theta }}$ .

## Step 2: Finding the maximum torque(a)

The maximum torque can be obtained when $\mathrm{\theta }=90°$ .

The area of the current-carrying loop is,

$\mathrm{A}=0.18\mathrm{m}×0.18\mathrm{m}\phantom{\rule{0ex}{0ex}}=0.03{\mathrm{m}}^{2}$

This can be used in equation (1) to determine the torque, such that,

${\mathrm{\pi }}_{\mathrm{max}}=\mathrm{NIABsin}\left(90°\right)\phantom{\rule{0ex}{0ex}}=150×50\mathrm{A}×0.03{\mathrm{m}}^{2}×1.60\mathrm{T}×\mathrm{sin}\left(90°\right)\phantom{\rule{0ex}{0ex}}=360\mathrm{N}‐\mathrm{m}$

Therefore, the maximum torque is $360\mathrm{N}‐\mathrm{m}$.

## Step 4: Finding the torque when the angle is (b)

Substitute the given data in equation (1), to determine the torque,

${\mathrm{\pi }}_{\mathrm{max}}=\mathrm{NIABsin}\left(90°\right)\phantom{\rule{0ex}{0ex}}=150×50\mathrm{A}×0.03{\mathrm{m}}^{2}×1.60\mathrm{T}×\mathrm{sin}\left(10.9°\right)\phantom{\rule{0ex}{0ex}}=68.1\mathrm{N}‐\mathrm{m}$

Hence, the value of torque is $68.1\mathrm{N}‐\mathrm{m}$.