A wire loop with 60 turns is formed into a square with sides of length s . The loop is in the presence of a 1.20 T uniform magnetic field B⃗ that points in the negative y direction. The plane of the loop is tilted off the x-axis by θ=15∘ . If i=2.50 A of current flows through the loop and the loop experiences a torque of magnitude 0.0186 N⋅m , what are the lengths of the sides s of the square loop, in centimeters?

the length of the sides s is  

Explanation:

From the question we are told that

     The number of turns is    

      The magnetic field is

     The angle the loop makes with the x-axis

      The current flowing through the loop is

       The magnitude of the torque is

        the length of the sides of the square is  

Generally, we can represent the torque magnitude as

Where A is the area of the square which is mathematically represented as

Substituting this into the formula for torque

making s the subject

    Substituting values

Converting to centimeters

2 cm

Explanation:

To fins the lengths of the sides of the loop you use the following formula for the calculation of the torque experienced by the loop in a magnetic field:

N: turns of the loop = 60

i: current in the loop = 2.50A

A: area of the loop = s*s

B: magnitude of the magnetic field = 1.20T

θ: angle between the plane of the lop and the direction of B = 15°

BY replacing the values of the torque and the other parameters in (1) you  can obtain the area of the loop:

but the area is s*s:

hence, the sides of the square loop have a length of 2cm

none, its a machine