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Q47PE

Expert-verifiedFound in: Page 813

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 ${\mathbf{0}}{\mathbf{.}}{\mathbf{650}}{\mathbf{\times}}{{\mathbf{10}}}^{\mathbf{\u2010}\mathbf{15}}$ m in radius with a current by a circular of ${\mathbf{1}}{\mathbf{.}}{\mathbf{05}}{\mathbf{\times}}{{\mathbf{10}}}^{{\mathbf{4}}}$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 $3.49\times {10}^{\u201025}\mathrm{N}\u2010\mathrm{M}$.

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.

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.

The area is given by,

$\mathrm{A}={\mathrm{\pi r}}^{2}\phantom{\rule{0ex}{0ex}}=\mathrm{\pi}\times {\left(0.650\times {10}^{\u201015}\mathrm{m}\right)}^{2}\phantom{\rule{0ex}{0ex}}=1.33\times {10}^{\u201030}{\mathrm{m}}^{2}$

The formula for the maximum torque is,

$\mathrm{\tau}=\mathrm{IAB}\phantom{\rule{0ex}{0ex}}=\left(1.05\times {10}^{4}\mathrm{A}\right)\times \left(1.33\times {10}^{\u201030}{\mathrm{M}}^{2}\right)\times 2.50\mathrm{T}\phantom{\rule{0ex}{0ex}}=3.49\times {10}^{\u201026}\mathrm{N}\u2010\mathrm{m}$

Therefore, the maximum torque on a proton is $3.49\times {10}^{\u201026}\mathrm{N}\u2010\mathrm{M}$

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