Question

A square current loop 5.5 cm on each side carries a 550 mA current. The loop is in a 1.1 T uniform magnetic field. The axis of the loop, perpendicular to the plane of the loop, is 30∘ away from the field direction. Part A What is the magnitude of the torque on the current loop?

Answer 1

Answer:

$\tau =0.000915\ Nm$

Explanation:

given,                        

side of loop = 5.5 cm = 0.055 m

current on each side of loop =  550 m A = 0.55 A

magnetic field on the loop = 1.1 T

axis of the loop make an angle of = 30°

torque = ?

$\tau = i \times A\times B \times sin \theta$

$\tau = 0.55 \times 0.055^2\times 1.1 \times sin 30^0$

$\tau =0.000915\ Nm$

hence, the torque on the current loop is $\tau =0.000915\ Nm$

Answer 2

The magnitude of the torque on the current loop is 9.15 x 10⁻⁴Am².

The given parameters;

• length of each side of the square, L = 5.5 cm = 0.055 m
• current in the coil, I = 550 mA
• magnetic field, B = 1.1 T
• direction of the field, θ = 30⁰

The area of the square loop is calculated as follows;

$A = l^2 = 0.055 \times 0.055\\\\A = 0.003025 \ m^2$

The magnitude of the torque on the current loop is calculated as follows;

$\tau = B \times I \times A \times sin(\theta)\\\\\tau = 1.1 \times 0.55 \times 0.003025 \times sin(30)\\\\\tau = 0.000915\\\\\tau = 9.15 \times 10^{-4} \ Am^2$

Thus, the magnitude of the torque on the current loop is 9.15 x 10⁻⁴Am².

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