A loop in which a constant current strength I = 15 A is maintained is freely established in a uniform magnetic

A loop in which a constant current strength I = 15 A is maintained is freely established in a uniform magnetic field (B = 0.025 T). The diameter of the coil is d = 20 cm. What work A must be done in order to rotate the coil about the axis coinciding with the diameter by the angle α = π?

To find the work necessary to rotate the loop under consideration, we use the formula: A = 0α∫М dα = 0α∫Рm * B * sinα dα = -I * S * B * cosα – 0 = -I * Π * d ^ 2/4 * B * cosα.

Variables: I – current in the considered loop (I = 15 A); d – coil diameter (d = 20 cm = 0.2 m); B – field induction (B = 0.025 T); α – turn angle of the coil (α = Π = 180º).

Let’s make a calculation: A = -I * Π * d ^ 2/4 * B * cosα = -15 * 3.14 * 0.2 ^ 2/4 * 0.025 * cos 180º = 0.011775 J ≈ 11.8 mJ.

Answer: To turn the loop in question, you need to do work of 11.8 mJ.