Question

A long electric cable is suspended above the earth and carries a current of 345 A parallel to the surface of the earth. The earth’s magnetic field has a value of 5.6 × 10−5 T and makes an angle of 88.2 ◦ with respect to the cable. What is the magnitude of the force f exerted on a cable of length 46.9 m due to the earth’s magnetic field? Answer in units of N

Answer 1

Answer:

0.906 N

Explanation:

Formula for magnetic force acting on current carrying cable:

$F = IBLsin(\theta)$

Where I = 345A is the current in the wire, B = $5.6*10^{-5} T$ is the magnetic magnitude generated by Earth. L = 46.9 m is the cable length. $\theta = 88.2^o$ is the angle between vector B and cable direction.

$F = 345*5.6*10^{-5}*46.9*sin(88.2^o)$

$F = 0.906 N$