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

Question

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aristeus aristeus · Answer 1 · 2020-04-20T06:20:15+02:00

Answer:

Force on wire is [tex]6.39\times 10^{-4}N[/tex]

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 [tex]n=25turns/cm=2500turns/m[/tex]

Magnetic field due to solenoid [tex]B=\mu ni[/tex]

[tex]B=4\pi \times 10^{-7}\times 2500\times 17=5.33\times 10^{-2}T[/tex]

Now magnetic force on the wire

[tex]F=iBl=0.150\times 5.33\times 10^{-2}\times 0.08=6.39\times 10^{-4}N[/tex]

So force on wire is [tex]6.39\times 10^{-4}N[/tex]