Question

A straight wire that is 25𝑐𝑚 long is carrying a current perpendicular to a magnetic field of 5.0 𝑇 and experiences a force of 8.0 × 10—3𝑁. What is the current in the wire?

Answer 1

We can find the current in the wire from the equation:

$$F_B=BILsin\theta,$$ $$I=\dfrac{F_B}{BLsin\theta},$$ $$I=\dfrac{8.0\cdot10^{-3}\ N}{5.0\ T\cdot0.25\ m\cdot sin90^{\circ}}=6.4\cdot10^{-3}\ A.$$