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### College Physics (Urone)

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

# A proton has a magnetic field due to its spin on its axis. The field is similar to that created by a circular to that created by a circular current loop ${\mathbf{0}}{\mathbf{.}}{\mathbf{650}}{\mathbf{×}}{{\mathbf{10}}}^{\mathbf{‐}\mathbf{15}}$ m in radius with a current by a circular of ${\mathbf{1}}{\mathbf{.}}{\mathbf{05}}{\mathbf{×}}{{\mathbf{10}}}^{{\mathbf{4}}}$A (no kidding ). Find maximum torque on a proton in a 2.50-T field. (This is a significant torque on a small particle.)

The maximum torque on a proton is $3.49×{10}^{‐25}\mathrm{N}‐\mathrm{M}$.

See the step by step solution

## Step 1:Definition of torque

Torque can be defined as the force acting on a current-carrying loop in a uniform magnetic field, which helps to rotate it in the circular orbit.

## Step 2: Finding the torque

First of all, we will get the area of the circular loop then we will use the equation $\mathrm{\tau }=\mathrm{IAB}$ to get the maximum torque for the angle to be 90o.

## Step 3: Finding the maximum torque

The area is given by,

$\mathrm{A}={\mathrm{\pi r}}^{2}\phantom{\rule{0ex}{0ex}}=\mathrm{\pi }×{\left(0.650×{10}^{‐15}\mathrm{m}\right)}^{2}\phantom{\rule{0ex}{0ex}}=1.33×{10}^{‐30}{\mathrm{m}}^{2}$

The formula for the maximum torque is,

$\mathrm{\tau }=\mathrm{IAB}\phantom{\rule{0ex}{0ex}}=\left(1.05×{10}^{4}\mathrm{A}\right)×\left(1.33×{10}^{‐30}{\mathrm{M}}^{2}\right)×2.50\mathrm{T}\phantom{\rule{0ex}{0ex}}=3.49×{10}^{‐26}\mathrm{N}‐\mathrm{m}$

Therefore, the maximum torque on a proton is $3.49×{10}^{‐26}\mathrm{N}‐\mathrm{M}$