Question

A non-uniform magnetic field exerts a net force on a current loop of radius R. FIGURE P29.78 shows a magnetic field that is diverging from the end of a bar magnet. The magnetic field at the position of the current loop makes an angle u with respect to the vertical. a. Find an expression for the net magnetic force on the current. b. Calculate the force if R = 2.0 cm, I = 0.50 A, B = 200 mT, and u = 20°.

Short answer

Part a

The expression for the net magnetic force on the current is $F_{net}=2\mathrm{\pi RIBsin\theta }$.

Part b

The force is$4.3\times {10}^{-3}N$.

Step-by-step solution

Given information

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

Part a

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 $F_y=\left(IB∆s\right)\sin \theta$.
• Thus the net force on the current loop is

role="math" localid="1650351977110" $$\begin{gathered} F_{Net}=\sum _{}{F}_y \\ =\sum _{}\left(IB∆s\right)\sin \theta \\ =IB\sin \theta \sum _{}∆s \end{gathered}$$

The sum of $∆s$is equal to the circumference of the loop, that is, data-custom-editor="chemistry" $2\mathrm{\pi aR}$.

Therefore, the net force is$F_{net}=2\mathrm{\pi RIBsin\theta }$.

Part b

Substitute the given values into the equation obtained in step 2

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

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