Neuwiedlq8

2023-02-01

A bar magnet has a magnetic moment of one joule per tesla. We need an external torque of 3 Nm to hold a bar magnet that makes 30 degrees with the field. Find the work required to turn it by further 90 degrees.

A) 6.3 J

B) 8.1 J

C) 4.9 J

D) 19.5 J

A) 6.3 J

B) 8.1 J

C) 4.9 J

D) 19.5 J

mystarynighthb5

Beginner2023-02-02Added 4 answers

The correct ANSWER is B. 8.1 J

(τ| = (M×B| = MBsinθ

$=MB\mathrm{sin}{30}^{\circ}=\frac{MB}{2}=3$

${W}_{ext}={U}_{f}-{U}_{i}=-MB\mathrm{cos}{\theta}_{f}+MB\mathrm{cos}{\theta}_{i}=6(\mathrm{cos}{30}^{\circ}-\mathrm{cos}{120}^{\circ})\phantom{\rule{0ex}{0ex}}=3(1+\sqrt{3})=3(2.7)=8.1J$

(τ| = (M×B| = MBsinθ

$=MB\mathrm{sin}{30}^{\circ}=\frac{MB}{2}=3$

${W}_{ext}={U}_{f}-{U}_{i}=-MB\mathrm{cos}{\theta}_{f}+MB\mathrm{cos}{\theta}_{i}=6(\mathrm{cos}{30}^{\circ}-\mathrm{cos}{120}^{\circ})\phantom{\rule{0ex}{0ex}}=3(1+\sqrt{3})=3(2.7)=8.1J$

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