Question

A circular circuit of radius 30 cm is built to rotate about a

pair of insulating posts. The circuit consists of a 100 volt

battery, two 25 kW resistors in parallel and one 75 kW

resistor in series. The circuit is exposed to an external uniform magnetic field

of magnitude 0.75 Tesla and directed 20 degrees from the

normal vector of the circuit and the axis of rotation.

Calculate the total torque (magnitude and direction) on the

circuit

Calculate the component of the torque on the circuit

(magnitude and direction) at point A (exactly ¼ of the way

around the loop)

Calculate the component of the torque on the circuit

(magnitude and direction) at point B (exactly 7/8 of the way

around the loop)

Answer 1

The torque is given by,

T = IA x B....(i) , where,

T = torque , I = current, A = area of cross-section, B = magnetic field and "x" denotes the cross-product.

Thus applying formula of cross-product, we have,

T = BIA sin (θ)....(ii), where θ = angle between normal vector and axis of rotation

Given data :

Radius = r = 30 cm = 0.3 m

Voltage = V_o = 100 V

Total power = P = (2 * 25) + 75 = 125 kW

Thus, effective resistance = R_eff = (V_o² / P) = (100² / 125 * 10³) Ω = 0.08 Ω

Thus, by Ohm's law effective current = I = V_o / R_eff = 100/0.08 = 1250 A

Magnetic field = B = 0.75 Tesla

Angle between normal vector and axis of rotation = θ = 20 degrees

Thus, from (ii), total torque

T_total = 0.75 * 1250 * (3.14 * 0.3²) * sin(20 degrees)

Or, T_total = 90.614 N-m (approx)

Thus, total torque on the circuit is approximately 90.614 N-m, directed vertically upward along the magnetic field, at 20 degrees with the normal plane of paper.

Exactly at (1/4)th the way around the loop, the value of angle between normal vector and axis of rotation = θ_1/4 = (20 + 90) degrees, since a circular angle is 360 degrees.

Thus, θ_1/4 = 110 degrees

Thus, torque = T_1/4 = 0.75 * 1250 * (3.14 * 0.3²) * sin(110 degrees) = 248.959 N-m (approx)

Thus assuming anti-clockwise rotation at (1/4)th the way around the loop, torque is approximately 248.959 N-m, directed along the magnetic field at 110 degrees with the normal plane of paper.

At (7/8)th the way around the loop, the value of angle between normal vector and axis of rotation = θ_7/8 = (20 + [90*0.5*7]) degrees

Or, θ_7/8 = 335 degrees

Thus, torque = T_7/8 = 0.75 * 1250 * (3.14 * 0.3²) * sin(335 degrees) = -111.967 N-m (approx)

Thus assuming anti-clockwise rotation at (7/8)th the way around the loop, torque is approximately -111.967 N-m, directed along the magnetic field at (360-335) = 25 degrees with the normal plane of paper.