A circular coil has moment of inertia 0.8 kg m2 around any diameter and is carrying current to produce a magnetic moment of 20 Am2 . The coil is kept initially in a vertical position and it can rotate freely around a horizontal diameter. When a uniform magnetic field of 4 T is applied along the vertical,it starts rotating around its horizontal diameter. The angular speed the coil acquires after rotating by 60o will be:
Solution
By energy conservation <br><br>U<sub>i</sub> + K<sub>i</sub> = U<sub>f</sub> + K<sub>f</sub><br><br>$\Rightarrow$ $- MB\,\cos 90^\circ + 0 = - MB\,\cos 30^\circ + {1 \over 2}I{\omega ^2}$<br><br>$\Rightarrow$ $MB{{\sqrt 3 } \over 2}$ $= {1 \over 2}I{\omega ^2}$<br><br>$\Rightarrow$ $\omega =$$\sqrt {{{MB\sqrt 3 } \over I}}$
<br><br>= $\sqrt {{{20 \times 4 \times \sqrt 3 } \over {0.8}}}$ = $10\sqrt {\sqrt 3 } = 10{\left( 3 \right)^{1/4}}$
About this question
Subject: Physics · Chapter: Magnetic Effects of Current · Topic: Biot-Savart Law
This question is part of PrepWiser's free JEE Main question bank. 96 more solved questions on Magnetic Effects of Current are available — start with the harder ones if your accuracy is >70%.