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4.24(d) A uniform magnetic field of 3000 G is established along the positive z-direction. A rectangular loop of sides 10 cm and 5 cm carries a current of 12 A. What is the torque on the loop in the given case shown in Figure (d)? What is the force in this case?

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The magnetic field is given as,

 \vec{B}=3000\ G \hat{k}=0.3\ T\hat{k}

Current in the loop = 12 Amp

Area of the loop = length\times breadth

The area under the loop (A)=0.1 \times 0.05 =0.005 m^2

\vec{A}=0.005\ m^{2}(-\frac{\hat{i}}{2}+\frac{\sqrt{3}}{2}\hat{j})

Torque is given by,

\\\vec{\tau }=I\vec{A}\times \vec{B}\\ \\\vec{\tau }=12\times 0.005(-\frac{\hat{i}}{2}+\frac{\sqrt{3}}{2}\hat{j})\times 0.3\hat{k}\\ \\\vec{\tau }=-0.018(\frac{\hat{i}}{2}+\frac{\sqrt{3}}{2}\hat{j})

The torque on the loop has a magnitude of 0.018 Nm and is at an angle of 240o from the positive-x direction. The force on the loop is zero.

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