If i = 1.70 A of current flows through the loop and the loop experiences a torque of magnitude 0.0760 N ⋅ m , what are the lengths of the sides s of the square loop, in centimeters?

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Question:

A wire loop with 40 turns is formed into a square with sides of length s. The loop is in the presence of a 2.0 T uniform magnetic field B that points in the negative y direction. The plane of the loop is tilted off the x-axis by 15°. If 1.70 A of current flows through the loop and the loop experiences a torque of magnitude 0.0760 N.m, what are the lengths of the sides s of the square loop, in centimeters?

Given Information:

Number of turns = N = 40 turns

Torque = τ = 0.0760 N.m

Current = I = 1.70 A

Magnetic field = B =  2 T

θ = 15°

Required Information:

Length of the sides of square loop = s = ?

s = 4.64 cm

Explanation:

τ = NIABsin(θ)

Where N is the number of turns, I is the current flowing through the square loop, A is the area of square loop, B is the magnetic field, and θ is the angle between square loop and magnetic field strength with respect to the x-axis.

Re-arranging the equation to find out A

A = τ/ NIBsin(θ)

A = 0.0760/40*1.70*2*sin(15)

A = 0.00215 m²

We know that area of a square is

A = s²

Taking square root on both sides yields

s = √A

s = √0.00215 = 0.0464 m

in centimeters

s = 4.64 cm

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