Question

An 8.00-cm-long piece of wire is formed into a square, and carries a clockwise current of 0.150 A. The loop is placed inside a solenoid, and the plane of the loop is perpendicular to the solenoid’s magnetic field. The solenoid carries a counterclockwise current of 17.0 A, and has 25 turns per centimeter. What is the force on each side of the loop?

A) 6.41E–4 N
B) 1.60E–4 N
C) 1.07E–5 N
D) 4.27E–3 N

Answer 1

Force on wire is
`6.39* 10^(-4)N`

Step-by-step explanation:

We have given length of wire l = 8 cm = 0.08 m

Current in the wire i = 0.150 A

Number of turns in solenoid
`n=25turns/cm=2500turns/m`

Magnetic field due to solenoid
`B=\mu ni`

`B=4\pi * 10^(-7)* 2500* 17=5.33* 10^(-2)T`

Now magnetic force on the wire

`F=iBl=0.150* 5.33* 10^(-2)* 0.08=6.39* 10^(-4)N`

So force on wire is
`6.39* 10^(-4)N`