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.

τ_max = 1.406 N.m

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

pe = weightxheight = 2000x1000=2,000,000joules