Question

A straight wire 15 cm long is in a uniform field of 0.4 T. This wire carries a current of 6 A. What is the force on the wire when it is at 30 degrees to the field?

Answer 1

The magnitude of the magnetic force produced on a current carrying wire with current I and length L which is placed in a region with a magnetic field B that forms an angle θ with the wire is given by:

`F=ILB\sin\theta`

Replace I=6A, L=0.15m, B=0.4T and θ=30º to find the magnitude of the force on the wire:

`F=(6A)(0.15m)(0.4T)\sin(30º)=0.18N`

Therefore, magnitude of the force on the wire is equal to 0.18N.