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A square current loop 4.90 cm on each side carries a 510mA current. The loop is in a 1.40 T uniform magneticfield. The axis of the loop, perpendicular to the plane of theloop, is 30 degrees away from the field direction.
What is the magnitude of the torque on the current loop?

Respuesta :

Olajidey

Answer:

the torque on the current loop is 0.000857Nm

Explanation:

Given that                  

side of loop = 4.90 cm = 0.049 m

current on each side of loop =  510 m A = 0.51 A

magnetic field on the loop = 1.40 T

axis of the loop make an angle of = 30°

torque = ?

T = i x A x B x sin(30)

[tex]\tau = i \times A\times B \times sin \theta\tau = 0.51 \times 0.049^2\times 1.4 \times sin 30^0\tau =0.000857\ Nm[/tex]

the torque on the current loop is 0.000857Nm

asuwafoh

Answer:

0.000857 Nm

Explanation:

From the question,

The formula of the torque on the current loop is given as

T = BIAsinФ..................... Equation 1

Where T = Torque, B = magnetic Field, I = current carried by the loop,  A = Area of the loop, Ф = direction of the magnetic field.

Given: B = 1.4 T, I = 510mA = 0.51 A, Ф = 30°

Since the loop is a square,

A = l²  Where l =  length of each side of the loop.

l = 4.9 cm = 0.049 m

A = 0.049²

A = 0.002401 m².

Substitute into equation 2

T = 1.4(0.51)(0.002401)(sin30°)

T = 1.4×0.51×0.002401×0.5

T = 0.000857 Nm