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

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weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

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so it requires 75 n force to move it upwards