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Q42PE

Expert-verifiedFound in: Page 813

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 $360\mathrm{N}\u2010\mathrm{m}$.

(b) When $\mathrm{\theta}=10.9\xb0$ , the torque equals $68.1\mathrm{N}\u2010\mathrm{m}$.

**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, ${\mathbf{\theta}}$ .**

The maximum torque can be obtained when $\mathrm{\theta}=90\xb0$ .

The area of the current-carrying loop is,

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

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

${\mathrm{\pi}}_{\mathrm{max}}=\mathrm{NIABsin}\left(90\xb0\right)\phantom{\rule{0ex}{0ex}}=150\times 50\mathrm{A}\times 0.03{\mathrm{m}}^{2}\times 1.60\mathrm{T}\times \mathrm{sin}\left(90\xb0\right)\phantom{\rule{0ex}{0ex}}=360\mathrm{N}\u2010\mathrm{m}$

Therefore, the maximum torque is $360\mathrm{N}\u2010\mathrm{m}$.

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

${\mathrm{\pi}}_{\mathrm{max}}=\mathrm{NIABsin}\left(90\xb0\right)\phantom{\rule{0ex}{0ex}}=150\times 50\mathrm{A}\times 0.03{\mathrm{m}}^{2}\times 1.60\mathrm{T}\times \mathrm{sin}\left(10.9\xb0\right)\phantom{\rule{0ex}{0ex}}=68.1\mathrm{N}\u2010\mathrm{m}$

Hence, the value of torque is $68.1\mathrm{N}\u2010\mathrm{m}$.

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