Question

A wire loop with 3030 turns is formed into a square with sides of length ss . The loop is in the presence of a 1.20 T1.20 T uniform magnetic field B⃗ B→ that points in the negative yy direction. The plane of the loop is tilted off the x-axisx-axis by θ=15∘θ=15∘ . If i=1.10 Ai=1.10 A of current flows through the loop and the loop experiences a torque of magnitude 0.0256 N⋅m0.0256 N⋅m , what are the lengths of the sides ss of the square loop, in centimeters?

Answer 1

2.59 cm

Step-by-step explanation:

The torque τ on a current carrying loop of wire is given by τ = NiABsinθ where N = number of turns of loop, i = current in loop, A = area of loop and B = magnetic field.

Now, given that τ = 0.0256 Nm, i = 1.10 A, B = 1.20 T,N = 30 and since the loop is tilted 15° off the x-axis and the magnetic field points in the negative y- direction, the angle between the normal to the loop and the magnetic field is thus 90° - 15° = 75°. So, θ = 75°.

We now find the area of the loop A from

τ = NiABsinθ

A = τ/NiBsinθ

substituting the values of the variables, we have

A = 0.0256 Nm/30 × 1.10 A × 1.20 T × sin75°

A = 0.0256 Nm/38.25

A = 6.69 × 10⁻⁴ m²

Since the loop is a square, with length of side L, its area A = L² and

L = √A

= √(6.69 × 10⁻⁴ m²)

= 2.59 × 10⁻² m

converting to cm, we have

L = 2.59 × 10⁻² m × 100 cm/m

L = 2.59 cm

So, the lengths of sides of the loop is 2.59 cm