91Ó°ÊÓ

$$\begin{gathered} R=2.0cm \\ I=0.50A \\ B=200mT \\ \mathrm{θ}=20° \end{gathered}$$

Consider the diagram below

• Consider a small segment of the loop of length Δs.
• The magnetic force on this segment is perpendicular to the current and the magnetic field.
• The figure shows two segments on opposite sides of the loop.
• The horizontal components of the forces cancel but the vertical components combine to give a force toward the bar magnet.
• The net force is the sum of the vertical components of the force on all segments around the loop. For a segment of length $∆s$, the magnetic force is $F=IB∆s$and the vertical component of the force is $F_y=\left(IB∆s\right)\sin \mathrm{θ}$.
• Thus the net force on the current loop is

role="math" localid="1650351977110" $$\begin{gathered} F_{Net}=\underset{}{∑}{F}_y \\ =\underset{}{∑}\left(IB∆s\right)\sin \mathrm{θ} \\ =IB\sin \mathrm{θ}\underset{}{∑}∆s \end{gathered}$$

The sum of $∆s$is equal to the circumference of the loop, that is, data-custom-editor="chemistry" $2\mathrm{Ï€²¹¸é}$.

Therefore, the net force is$F_{net}=2\mathrm{Ï€¸é\pm õ\mu þ²õ¾\pm ²Ôθ}$.

Substitute the given values into the equation obtained in step 2

$$\begin{gathered} F_{net}=2\mathrm{π}\left(0.02\right)×0.50×200×{10}^{-3}×\sin 20° \\ =4.3×{10}^{-3}\mathrm{N} \end{gathered}$$

Therefore, the force is$4.3×{10}^{-3}N$.