Question:
What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30º with the direction of a uniform magnetic field of 0.15 T?
Solution:
Current in the wire, I = 8 A
Magnitude of the uniform magnetic field, B = 0.15 T
Angle between the wire and magnetic field, θ = 30°.
Magnetic force per unit length on the wire is given as:
$f=B / \sin \theta$
$=0.15 \times 8 \times 1 \times \sin 30^{\circ}$
$=0.6 \mathrm{~N} \mathrm{~m}^{-1}$
Hence, the magnetic force per unit length on the wire is 0.6 N m–1.