A solid cylinder of radius R has total charge Q distributed uniformly over its volume. It is rotating about its axis with angular speed ω. The magnitude of the total magnetic moment of the cylinder is (A) QR^2ω (B) 1/2 QR^2ω (C) 1/4 QR^2ω (D) 1/8 QR^2ωA solid cylinder of radius R has total charge Q distributed uniformly over its volume. It is rotating about its axis with angular speed ω. The magnitude of the total magnetic moment of the cylinder is (A) QR^2ω (B) 1/2 QR^2ω (C) 1/4 QR^2ω (D) 1/8 QR^2ω

To determine the magnetic moment of a rotating charged solid cylinder, we need to consider how the charge distribution and rotation contribute to the magnetic moment. The magnetic moment (μ) is a vector quantity that represents the strength and direction of a magnetic source. For a rotating charged object, the magnetic moment can be derived from the motion of the charges within the object.

Understanding the Magnetic Moment of a Rotating Cylinder

When a solid cylinder of radius R carries a total charge Q and rotates about its axis with an angular speed ω, each infinitesimal charge element contributes to the overall magnetic moment. The key steps to derive the total magnetic moment are as follows:

1. Charge Distribution

The total charge Q is uniformly distributed throughout the volume of the cylinder. The volume charge density (ρ) can be calculated as:

• ρ = Q / V
• Where V = πR²h (the volume of the cylinder, with height h).

2. Current Due to Rotation

As the cylinder rotates, the charge creates a current. The current (I) can be thought of as the charge passing through a point per unit time. For a rotating cylinder, the effective current can be expressed as:

• I = Q / T
• Where T is the period of rotation, given by T = 2π/ω.

Substituting T into the equation for current gives:

• I = Qω / (2π).

3. Magnetic Moment Calculation

The magnetic moment (μ) for a current loop is given by:

• μ = I * A
• Where A is the area of the loop. For our cylinder, we can consider the area of a circular cross-section, which is A = πR².

Substituting the expression for current into the magnetic moment formula yields:

• μ = (Qω / (2π)) * (πR²) = (QR²ω) / 2.

Final Result

Thus, the magnitude of the total magnetic moment of the rotating solid cylinder is:

• μ = (1/2) QR²ω.

From the options provided, the correct answer is (B) 1/2 QR²ω.

Summary

In summary, the magnetic moment of a uniformly charged rotating solid cylinder can be derived by considering the charge distribution, the effective current due to rotation, and the area associated with the charge. The final expression shows that the magnetic moment is directly proportional to the total charge, the square of the radius, and the angular speed of rotation.