Question

A 15-turn rectangular loop of wire of width 10 cm and length 20 cm has a current of 2.5 A flowing through it. Two sides of the loop are oriented parallel to a uniform magnetic field of strength 0.037 T, and the other two sides are perpendicular to the magnetic field. A)What is the magnitude of the magnetic moment of the loop?B)What torque does the magnetic field exert on the loop?

Answer 1

Given Information:

Magnetic field = B = 0.037 T

Current = I = 2.5 A

Number of turns = N = 15 turns

Length of rectangular coil = L = 20 cm = 0.20 m

Width of rectangular coil = W = 10 cm = 0.10 m

Required Information:

(a) Magnetic moment = µ = ?

(b) Torque = τ = ?

Answer:

(a) Magnetic moment = 0.75 A.m ²

(b) Torque = 0.0277 N.m

Explanation:

(a) The magnetic moment µ is given by

µ = NIA

Where µ is the magnetic, N is the number of turns, I is the current, A is the area of rectangular loop and is given by

A = W*L

A = 0.10*0.20

A = 0.02 m²

µ = 15*2.5*0.02

µ = 0.75 A.m²

(b) The toque τ exerted on current carrying loop with A area in the presence of a magnetic field B is given by

τ = NIAB

τ = 15*2.5*0.02*0.037

τ = 0.0277 N.m

Alternatively,

τ = µB

τ = 0.75*0.037

τ = 0.0277 N.m

Answer 2

(a) the magnitude of the magnetic moment of the loop is 0.75 Am²

(b) the torque the magnetic field exerts on the loop is 0.028 N.m

Step-by-step explanation:

Given;

number of turns, N = 15

width of the loop, w = 10 cm = 0.1 m

length of loop, L = 20 cm = 0.2 m

current through the loop, I = 2.5 A

strength of the magnetic field, B = 0.037 T

Area of the loop, A = L x w = 0.2 x 0.1 = 0.02 m²

Part (a) the magnitude of the magnetic moment of the loop

μ = NIA

where;

μ is the magnitude of the magnetic moment of the loop

μ = 15 x 2.5 x 0.02 = 0.75 Am²

Part (b) the torque the magnetic field exerted on the loop

τ = μB

where;

τ is the torque the magnetic field exerts on the loop

τ = μB = 0.75 x 0.037 = 0.028 N.m