Question

Assume that the loop is initially positioned at θ=30∘ and the current flowing into the loop is 0.500 A . If the magnitude of the magnetic field is 0.300 T , what is τ, the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field?

Answer 1

the net torque about the vertical axis of the current loop is,

$$ \tau=N I B A \sin \phi $$

Here, \(\phi\) is the angle between the normal (to the loop) and

the magnetic field.

$$ \phi=90^{\circ}-30^{\circ}=60^{\circ} $$

Hence, the net torque about the vertical axis of the current loop is,

$$ \begin{aligned} \tau &=(1) I B A \sin 60^{\circ} \\ &=I B A \sin 60^{\circ} \end{aligned} $$

If \(I=0.500 \mathrm{~A}, B=0.300 \mathrm{~T}, A=(0.04 \mathrm{~m})(0.02 \mathrm{~m})\), then the net

torque about the vertical axis of the current loop is,

$$ \begin{array}{l} \tau=(0.500 \mathrm{~A})(0.300 \mathrm{~T})(0.04 \mathrm{~m})(0.02 \mathrm{~m}) \sin 60^{\circ} \\ \tau \approx 0.000104 \mathrm{~N} \cdot \mathrm{m} \end{array} $$