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?

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 = ?

(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

volts times amps 6 x 0.5 3w

the maxium height is 34m/s