A 50-turn circular coil of radius 2.0 cm carrying a current of 5.0 A is rotated in a uniform magneti

Question

A 50-turn circular coil of radius 2.0 cm carrying a current of 5.0 A is rotated in a uniform magnetic field of strength 0.20 T . In a particular position of the coil, the torque acting on it is half of the maximum possible torque. If the acute angle between the magnetic field and the plane of the coil in this position is $	heta^{\circ}$, find the value of $	heta$.

Professor PC Thomas Classes · Answer

$$ \begin{gathered} 	au=\frac{1}{2} 	au_{\max } \ N I A B \sin \phi=\frac{1}{2} N I A B \ \sin \phi=\frac{1}{2} \ \phi=30^{\circ} \end{gathered} $$ (Note: $\phi$ is the angle between the normal and the magnetic field.) The question asks for the angle $	heta$ between the magnetic field and the plane of the coil. The plane of the coil is perpendicular to its normal vector, so: $$ \begin{gathered} 	heta=90^{\circ}-\phi \ 	heta=90^{\circ}-30^{\circ}=60^{\circ} \end{gathered} $$Therefore, the value of $	heta$ is $\mathbf{6 0}$.