A solenoid having 165 turns and a cross-sectional area of 6.75 cm^2 carries a current of 1.20 A.

step 1 So we have some givens. We know the formula for the torque on a solenoid is going to be the numb

A solenoid having 165 turns and a cross-sectional area of...

A solenoid having 165 turns and a cross-sectional area of 6.75 $\mathrm{cm}^{2}$ carries a current of 1.20 A. If it is placed in a uniform 1.12 T magnetic field, find the torque this field exerts on the solenoid if its axis is oriented (a) perpendicular to the field, (b) parallel to the field, (c) at 35. $0^{\circ}$ with the field.

Key Concepts

Angular Dependence of Magnetic Torque

The torque on a magnetic dipole depends on the relative orientation between the dipole moment and the magnetic field, expressed as sin(?). When ? is 90° (dipole perpendicular to the field), the torque is maximized because sin(90°) = 1. Conversely, when ? is 0° or 180° (dipole parallel or antiparallel to the field), the sin(?) factor becomes zero, resulting in no torque. This dependence is crucial in understanding rotational equilibrium and alignment in magnetic systems.

Magnetic Dipole Moment

A magnetic dipole moment is produced by a current loop or coil and is defined as ? = N * I * A, where N is the number of turns in the loop, I is the current, and A is the cross-sectional area of the loop. It represents the strength and orientation of the magnetic source and is a fundamental quantity when analyzing how current-carrying objects interact with external magnetic fields.

Torque on a Magnetic Dipole

When a magnetic dipole is placed in a uniform magnetic field, it experiences a torque given by ? = ? × B, with a magnitude of ? = ? * B * sin(?), where ? is the angle between the dipole moment and the magnetic field. This torque tends to rotate the dipole so that its magnetic moment aligns with the magnetic field, and its effectiveness depends strongly on the sine of the angle between the two.

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