Magnetic Force on a Current-Carrying Wire — Question 3

Question 3

Suspension angle measures a horizontal magnetic force on a current-carrying wire

A wire of length L=.200 m and mass .0500 kg carries I=4.00 A along +z. A uniform field $+.300\hat y$ T acts on it. Two identical light suspension strings tilt by angle theta from vertical. The return leads exert negligible force. Use g=9.81 m/s squared.

1. Find the magnetic force magnitude and string angle.

2. Find the tension in each string and its change if the current is reversed.

Question 3 - Solution

Suspension angle measures a horizontal magnetic force on a current-carrying wire

(a) Method and calculation

$$\vec F_B=IL\hat z\times B\hat y=-ILB\hat x,\quad F_B=.240\,\mathrm N.$$$$\tan\theta=F_B/(mg)=.240/.4905\Rightarrow\theta=\tan^{-1}(.240/.4905).$$

(b) Interpretation and further calculation

$$2T\cos\theta=mg\Rightarrow T=\tfrac12\sqrt{(mg)^2+F_B^2}=.273034\,\mathrm N.$$$$I\to-I:\quad\theta\to-\theta,\quad T\text{ unchanged}.$$

Physical check

The sign of the horizontal displacement changes under current reversal, but the force magnitudes and tensions do not.