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.Determine the magnitude of the change in length of the spring when a magnetic field B = 2 T is applied along the vertical axis.

Question

Quizplus · Accepted Answer

The magnetic field will interact with the current flowing in the ring, creating a torque that will cause the ring to rotate. The direction of the torque can be determined using the right-hand rule. If the thumb of the right hand points in the direction of the magnetic field, and the fingers point in the direction of the current, then the palm will face the direction of the torque.

In this case, the magnetic field is pointing straight up and the current is flowing clockwise. Therefore, the torque will be counterclockwise, causing the ring to rotate.

The amount of torque can be calculated using the formula:

τ = NABsinθ

Where N is the number of turns (in this case, N = 1), A is the area of the ring (A = πr^2 = 16π cm^2), B is the magnetic field strength (B = 2 T), and θ is the angle between the magnetic field and the plane of the ring (θ = 90°).

Plugging in the values, we get:

τ = (1)(16π)(2)(sin 90°) 
   = 32π N·cm

This torque will cause the spring to compress. The amount of compression can be calculated using the formula for Hooke's law:

F = -kx

Where F is the force exerted by the spring, k is the spring constant (k = 100 N/m), and x is the displacement of the spring from its natural length.

The force exerted by the spring is equal and opposite to the torque:

F = -τ

Plugging in the value for τ, we get:

F = -32π N·cm

This force will cause the spring to compress by an amount x:

x = -F/k

Plugging in the values for F and k, we get:

x = -(-32π N·cm)/(100 N/m) 
   = 0.32π m

This is equivalent to 3.2 cm. However, we need to convert this to the change in length of the spring:

ΔL = 2x

Because the spring will compress by a distance x on one side, and extend by the same amount on the other side, the total change in length will be twice x.

Plugging in the value for x, we get:

ΔL = 2(0.32π m) 
     = 2(3.2 cm) 
     = 6.4 cm

Therefore, the magnitude of the change in length of the spring is 6.4 cm, which is closest to option C (2.5 mm).

The Figure Shown Depicts the Cross Section of a Current