JEE MAIN - Physics (2020 - 4th September Evening Slot - No. 19)
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:
10 $$\pi $$ rad s–1
20 $$\pi $$ rad s–1
$$10{\left( 3 \right)^{1/4}}$$ rad s–1
20 rad s–1
Explanation
By energy conservation
Ui + Ki = Uf + Kf
$$ \Rightarrow $$ $$ - MB\,\cos 90^\circ + 0 = - MB\,\cos 30^\circ + {1 \over 2}I{\omega ^2}$$
$$ \Rightarrow $$ $$MB{{\sqrt 3 } \over 2}$$ $$ = {1 \over 2}I{\omega ^2}$$
$$ \Rightarrow $$ $$\omega =$$$$\sqrt {{{MB\sqrt 3 } \over I}} $$
= $$\sqrt {{{20 \times 4 \times \sqrt 3 } \over {0.8}}} $$ = $$10\sqrt {\sqrt 3 } = 10{\left( 3 \right)^{1/4}}$$
Ui + Ki = Uf + Kf
$$ \Rightarrow $$ $$ - MB\,\cos 90^\circ + 0 = - MB\,\cos 30^\circ + {1 \over 2}I{\omega ^2}$$
$$ \Rightarrow $$ $$MB{{\sqrt 3 } \over 2}$$ $$ = {1 \over 2}I{\omega ^2}$$
$$ \Rightarrow $$ $$\omega =$$$$\sqrt {{{MB\sqrt 3 } \over I}} $$
= $$\sqrt {{{20 \times 4 \times \sqrt 3 } \over {0.8}}} $$ = $$10\sqrt {\sqrt 3 } = 10{\left( 3 \right)^{1/4}}$$
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