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  • A coil in the shape of an equilateral triangle of side 10 cm lies in a vertical plane between the pole pieces of permanent magnet producing a horizontal magnetic field 20 mT.The torque acting on the coil when a current of 0.2 A is passed through it and it

A coil in the shape of an equilateral triangle of side 10 cm lies in a vertical plane between the pole pieces of permanent magnet producing a horizontal magnetic field 20 mT.The torque acting on the coil when a current of 0.2 A is passed through it and its plane becomes parallel to the magnetic field will be \sqrt{x}\times 10^{-5}\: Nm. The value of x is_____.
 

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When the plane becomes parallel to magnetic field, then angle between Dipole moment & B is 90^{\circ}
\tau= MB\sin 90^{\circ}= MB
   = IAB
         
A= \frac{1}{2}\times b\times h= \frac{1}{2}\times l\times \frac{\sqrt{3}l}{2}
A= \frac{\sqrt{3}}{4}l^{2}= \frac{\sqrt{3}}{4}\times 10^{-2}
\tau = IAB= 0\cdot 2\times \frac{\sqrt{3}}{4}\times 10^{-2}\times 20\times 10^{-3}
\tau =\sqrt{3}\times 10^{-5}\, Nm
\therefore x= 3

Posted by

Ritika Jonwal

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