A conducting circular loop is rotated about its diameter at a constant angular speed of $100 \mathrm

Question

A conducting circular loop is rotated about its diameter at a constant angular speed of $100 \mathrm{rad} / \mathrm{s}$ in a magnetic field of 0.5 T perpendicular to the axis of rotation. When the loop is rotated by $30^{\circ}$ from the horizontal position, the induced EMF is 15.4 mV . The radius of the loop is $\_\_\_\_$ mm. $ \left(	ext { Take } \pi=\frac{22}{7}ight) $

Professor PC Thomas Classes · Answer

The magnetic flux is $\Phi = \vec{B} \cdot \vec{A} = BA \cos	heta$, where $	heta = \omega t$ is the angle between the normal to the area and the magnetic field. The induced EMF is $E = -\frac{d\Phi}{dt} = BA\omega \sin(\omega t)$. When the loop is $30^\circ$ from the horizontal, the angle with the field corresponds to $\omega t = 30^\circ$ (or $60^\circ$ depending on reference, but evaluating gives the matched magnitude). Substituting knowns: $15.4 	imes 10^{-3} = (0.5) (\frac{22}{7} r^2) (100) \sin(30^\circ)$. Thus, $15.4 	imes 10^{-3} = 25 	imes \frac{22}{7} r^2 = \frac{550}{7} r^2$. Solving for radius squared: $r^2 = \frac{15.4 	imes 10^{-3} 	imes 7}{550} = 196 	imes 10^{-6} 	ext{ m}^2$. Taking the square root gives $r = 14 	imes 10^{-3} 	ext{ m} = 14 	ext{ mm}$.