The question is incomplete! The complete question along with answer and explanation is provided below
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