Question 176: Magnetism and Matter

Question

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:

Accepted Answer

By energy conservation
U
i
+ K
i
= U
f
+ K
f
$$ \Rightarrow $$ $$ - MB\,\cos 90^\circ  + 0 =  - MB\,\cos 30^\circ  + {1 \over 2}I{\omega ^2}$$
$$ \Rightarrow $$ $$MB{{\sqrt 3 } \over 2}$$ $$ = {1 \over 2}I{\omega ^2}$$
$$ \Rightarrow $$ $$\omega  =$$$$\sqrt {{{MB\sqrt 3 } \over I}}   $$
= $$\sqrt {{{20 	imes 4 	imes \sqrt 3 } \over {0.8}}} $$ = $$10\sqrt {\sqrt 3 }  = 10{\left( 3 ight)^{1/4}}$$

Answer

10 $$\pi $$ rad s–1

Answer

20 $$\pi $$ rad s–1

Answer

$$10{\left( 3 \right)^{1/4}}$$ rad s–1

Answer

20 rad s–1

Magnetism and Matter

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