JEE MainPhysicsToughMagnetism

A 100 turns closely wound current carrying coilof radius 10 cm
is placed in a vertical plane. Thecoil carries a current of 3.2 A and
free to rotateabout a horizontal axis which coincides with itsdiameter.
A uniform horizontal magnetic field of2 Tesla exists in the region. Due
to magnetic torquecoil starts rotating. The angular velocity in
rad/secof the coilwhen it has rotated by $\frac{\pi}{2}$ is 10n. Find n.

Show answer and solution

\tau = MBsin\theta

I\alpha = MBsin\theta

\alpha = \frac{MB}{I}\sin\theta

\frac{d\omega}{d\theta}*\frac{d\theta}{dt} = \frac{MB}{I}sin\theta

\int_{0}^{\omega}\mspace{2mu}\omega d\omega = \int_{0}^{\frac{\pi}{2}}\mspace{2mu}\frac{MB}{I}\sin{\theta d\theta}

\left( \frac{\omega^{2}}{2} \right)_{0}^{\omega} = \frac{MB}{I}\lbrack - cos\theta\rbrack_{0}^{\pi/2}

\omega = \sqrt{\frac{2MB}{I}} = \sqrt{\frac{2NIAB}{I_{rot}}}\

= \sqrt{\frac{2 \times 100 \times 3.2 \times \pi \times 100 \times 10^{- 4} \times 2}{0.1}}

= 20 rad/sec.

Source: Grass-G / P-13 Physics paper 4 Nov.docx

Need a video for this idea?

Scan any doubt from your notes and get a whiteboard explanation.