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  • A square loop of mass M side 'a', and carrying current io lies in the xy plane as shown in the figure. It is hinged at y axis so that it can freely rotate about it. Mom

A square loop of mass M side 'a', and carrying current io lies in the xy plane as shown in the figure. It is hinged at y axis so that it can freely rotate about it. Moment of inertia of the loop about an axis through its c.m. and normal to its plane is equal to \gamma \mathrm{ma} \wedge 2. At \mathrm{t}=0 an external magnetic field of induction B is applied along +ve x axis. The initial angular acceleration of the loop is equal t

Option: 1

\frac{4 \mathrm{i}_0 \mathrm{~B}}{\mathrm{M}(2 \gamma+1)}


Option: 2

\frac{\mathrm{i}_0 \mathrm{~B}}{\mathrm{M} \gamma}


Option: 3

\frac{4 \mathrm{i}_{\mathrm{o}} \pi \mathrm{B}}{\mathrm{M}(2 \gamma+1)}


Option: 4

\frac{4 \mathrm{i}_{\circ} \mathrm{B}}{\mathrm{M}(2 \gamma+1)}


Answers (1)

best_answer

\mathrm{l \gamma =m \frac{q^2}{4}+\frac{\gamma m q^2}{2} \\ }

\mathrm{ =m \frac{q^2}{4}(1+2 \gamma) \\ }

\mathrm{ \tau =l q^2 B(\bar{m} \times \bar{B}) \\ }

\mathrm{ \alpha =\frac{\tau}{l \gamma}=\frac{4 I B}{m(2 \gamma+1)} }

Posted by

Gautam harsolia

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