Question

A 3.02 m long wire loop carrying a current of 2.25 A is in the shape of an equilateral triangle. If the loop is placed in a constant magnetic field of magnitude 0.475 T, determine the maximum torque that acts on it.

Answer 1

τ_max = 1.406 N.m

Step-by-step explanation:

We are given:

Perimeter of the equilateral triangle loop = 3.02 m

Current in the loop; I = 2.25 A

Magnetic field strength; B = 0.475 T

The maximum torque, is given by the formula:

τ_max = N•I•A•B

Now, N is 1 turn

B and I are given but we don't have the Are of the triangle A.

Now, we are given the perimeter of the equilateral triangle as 3.02m

Thus,the length of one side is; (3.02/3)m

Since all angles in an equilateral triangle are equal, we can find the height from trigonometric ratios as;

h/3.02 = sin 60

h = 3.02 x 0.866 = 2.615m

Area of triangle = (1/2)((3.02/3))(2.615) = 1.316 m²

Thus,

τ_max = N•I•A•B = 1•(2.25)(1.316)(0.475) = 1.406 N.m

Answer 2

0.78 Nm

Step-by-step explanation:

Parameters given:

Length of the loop, L = 3.02 m

Current in the loop, I = 2.25 A

Magnetic field strength, B = 0.475 T

The torque acting on a loop of wire with radius, r, carrying a current, I, in magnetic field, B, is given as:

τ = N * I * A * B

Where N = number of turns = 1

A = area of loop = pi * r²

We do not have the radius of the loop, but we can find it. The length of the loop is the same as the circumference. Hence, we can find the radius and the area of the loop.

L = 2 * pi * r

3.02 = 2 * pi * r

=> r = 0.481 m

The area, A, will be:

A = pi * r² = pi * 0 481² = 0.726 m²

Therefore, the torque acting on the loop of wire is:

τ = 0.726 * 2.25 * 0.475

τ = 0.78 Nm