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A rectangular loop of wire with a cross-sectional area of 0.339 m2 carries a current of 0.145 A. The loop is free to rotate about an axis that is perpendicular to a uniform magnetic field strength of 0.83 T. The plane of the loop is initially at an angle of 39.183o to the direction of the magnetic field. What is the magnitude of the torque on the loop ?



Answer :

Given:

The area of the loop is A = 0.339 m^2

The current in the loop is I = 0.145 A

The strength of the magnetic field is B = 0.83 T

The number of turns in the loop is n = 1

The angle is

[tex]\theta=\text{ 39.183}^{\circ}[/tex]

To find the torque on the loop.

Explanation:

The torque can be calculated by the formula

[tex]\tau=nIABsin\theta[/tex]

On substituting the values, the torque will be

[tex]\begin{gathered} \tau=1\times0.145\times0.339\times0.83\times sin(39.183^{\circ}) \\ =0.0258\text{ N m} \end{gathered}[/tex]

Thus, the magnitude of the torque on the loop is 0.0258 N m

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