The figure shown depicts the cross section of a current carrying ring of wire having a radius of 4 cm.A 1-A current flows in the ring in the direction indicated in the drawing.The ring is free to rotate,clockwise or counterclockwise,about an axis normal to the page which passes through its center.The ring is connected to a spring with a spring constant of 100 N/m as shown.The spring is fixed to a wall at the other end.When the ring is in the vertical position,the spring is at its natural length.One may assume that the length of the spring is long enough such that any extension or compression follows Hooke's law.If the magnitude of the change in length of the spring when a magnetic field B is applied along the vertical axis is 1.25 mm,what is the magnitude of the magnetic field?

Question

Quizplus · Accepted Answer

The torque on the ring due to the magnetic field is balanced by the torque due to the spring. The magnetic moment of the ring is given by the product of current and area (I * π * r^2). The torque due to the magnetic field is the product of magnetic moment and magnetic field (μ * B). The torque due to the spring is the product of spring constant, change in length, and radius (k * Δx * r). Equating these and solving for B gives B = (k * Δx) / (I * π * r) = (100 * 0.00125) / (1 * π * 0.04) = 1.0 T.

The Figure Shown Depicts the Cross Section of a Current