Magnetic Force on a Current-Carrying Wire — Question 1

Question 1

The directed length vector determines the force on a straight current segment

A straight segment carries I=3.00 A from its first endpoint to its second, with directed displacement $\vec L=(.200\hat x+.100\hat y)$ m. The uniform field is $\vec B=(.400\hat x-.300\hat z)$ T.

1. Find the force vector and magnitude.

2. Verify that the force is perpendicular to both the segment and the magnetic field.

Question 1 - Solution

The directed length vector determines the force on a straight current segment

(a) Method and calculation

$$\vec L\times\vec B=(-.0300\hat x+.0600\hat y-.0400\hat z)\,\mathrm{T\,m}.$$$$\vec F=I\vec L\times\vec B=(-.0900\hat x+.180\hat y-.120\hat z)\,\mathrm N,\quad F=\sqrt{.0549}\,\mathrm N.$$

(b) Interpretation and further calculation

$$\vec F\cdot\vec L=(-.0900)(.200)+(.180)(.100)=0.$$$$\vec F\cdot\vec B=(-.0900)(.400)+(-.120)(-.300)=0.$$

Physical check

The length vector follows conventional current, regardless of which microscopic charge carriers carry it.