A current loop, carrying a current of 5.6 A, is in the shape of a right triangle with sides 30, 40, and 50 cm. The loop is in a uniform magnetic field of magnitude 110 mT whose direction is parallel to the current in the 50 cm side of the loop. Find the magnitude of (a) the magnetic dipole moment of the loop in amperes-square meters and (b) the torque on the loop.

Answer:0.0336Am^2; 0.037Nm

Explanation:

Parameters:

Base of triangle (b) = 30cm = 0.3m

Height of triangle (h) = 40cm = 0.4m

Magnetic field magnitude (B) = 110mT = 0.11T

Current (I) = 5.6A

Magnetic dipole moment = current * Area

Area of triangle = 0.5*b*h

Area of triangle = 0.5*0.3*0.4 = 0.06m^2.

1.) Magnetic dipole moment = 5.6 * 0.06 = 0.336 Am^2

2.) Torque (T) = Magnetic dipole moment * B

T = 0.336 * 0.11 = 0.0369 = 0.037Nm

a) μ = 0.336 A. m²

b) T = 0.037 N. m

Explanation:

Given:

The current loop is carrying a current of 5,6 A in the shape of a right triangle with sides 30, 40 and 50 cm. From the sides of the triangle we can deduce that hypotenuse which is of a larger value is 50 cm while the 30 cm and 40 cm are for the base and height. The loop has an uniform magnetic field of magnitude 110 mT whose direction is parallel to the current in the 50 cm side of the loop.

(a) the magnetic dipole moment of the loop in amperes-square meters

Given:

Base (b) and height (H) = 30 cm and 40 cm = 0.3 m and 0.4 m

I = 5.6 A

Area of the triangle (A) = 1/2 (b * h)

A = 1/2 (b * h) = 1/2 (0.3 * 0.4) = 0.06 m²

Dipole moment (μ) = IA

μ = IA = 5.6 * 0.06 = 0.336 A. m²

μ = 0.336 A. m²

b) The torque on the loop.

Torque (T) = μB

B = 110 mT = 110 * 10⁻³ T

T = μB = 110 * 10⁻³ * 0.336 = 0.037 N. m

T = 0.037 N. m