Question

A circular loop of wire of area 10 cm^2 carries a current of 25 A. At a particular instant, the loop lies in the xy-plane and is subjected to a magnetic field B =(2.0iˆ+6.0jˆ+8.0kˆ)×10−3T. As viewed from above the xy-plane, the current is circulating clockwise. (a) What is the magnetic dipole moment of the current loop? (b) At this instant, what is the magnetic torque on the loop?

Answer 1

The magnetic dipole moment of the current loop is  0.025 Am².

The magnetic torque on the loop is 2.5 x 10⁻⁴  Nm.

What is magnetic dipole moment?

The magnetic dipole moment of an object, is the measure of the object's tendency to align with a magnetic field.

Mathematically, magnetic dipole moment is given as;

μ = NIA

where;

• N is number of turns of the loop
• A is the area of the loop
• I is the current flowing in the loop

μ = (1) x (25 A) x (0.001 m²)

μ = 0.025 Am²

The magnetic torque on the loop is calculated as follows;

τ = μB

where;

• B is magnetic field strength

B = √(0.002² + 0.006² + 0.008²)

B = 0.01 T

τ = μB

τ =  0.025 Am² x 0.01 T

τ = 2.5 x 10⁻⁴  Nm

Thus, the magnetic dipole moment of the current loop is determined from the current and area of the loop while the magnetic torque on the loop is determined from the magnetic dipole moment.

Learn more about magnetic dipole moment here: brainly.com/question/13068184

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