Question

What is the magnitude of the magnetic force per unit length on a wire carrying a current of 8 A making an angle of `30^@` with the direction of a uniform magnetic field of `0*15T`?

Answer 1

Current in the wire, `I = 8A`
Magnitude of the uniform magnetic field, `B = 0.15T`
Angle between the wire and magnetic field, `theta = 30^(@)`.
Magnetic force per unit length on the wire is given as:
`f = BI sin theta`
`=0.15 xx 8 xx1 xx sin 30^(@)`
`= 0.6 N m^(-1)`
Hence, the magnetic force per unit length on the wire is `0.6 Nm^(-1)`