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  • A rectangular coil of length 0.12 m and width 0.1 m having 50 turns of wire is suspended vertically in a uniform magnetic field of strength  <img alt="\mathrm{0.2 weber/m^{2}}" src="https

A rectangular coil of length 0.12 m and width 0.1 m having 50 turns of wire is suspended vertically in a uniform magnetic field of strength  \mathrm{0.2 weber/m^{2}}  . The coil carries a current of 2A. If the plane of the coil is inclined at an angle of 30° with the direction of the field, the torque required to keep the coil in stable equilibrium will be :

Option: 1

0.20 Nm


Option: 2

0.24 Nm


Option: 3

0.12 Nm


Option: 4

0.15 Nm


Answers (1)

best_answer

Here, number of turns of coil,\mathrm{N}=50 
Current through the coil  \mathrm{I}=2 \mathrm{~A}
Area \mathrm{A=I \times b=0.12 \times 0.1 \mathrm{~m} 2}
Magnetic field \mathrm{^{\overrightarrow{ }}=0.2 \mathrm{\omega} / \mathrm{m}^{\wedge} 2} 

Torque required to keep the coil in stable equilibrium.

\mathrm{ \tau=\overrightarrow{\mathrm{M}} \times \overrightarrow{\mathrm{B}}=\mathrm{MB} \sin 60^{\circ}=\mathrm{Ni} \mathrm{AB} \sin 60^{\circ} \\ }

\mathrm{ =50 \times 2 \times 0.12 \times 0.1 \times 0.2 \times \frac{\sqrt{3}}{2} \\ }

\mathrm{ =12 \sqrt{3} \times 10^{-2}=0.20784 \mathrm{Nm} }.
$$

 

Posted by

Rishabh

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