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The radius of a circular coil having 100 turns is 5 cm and a current of 0.5A is flowing through this coil. If it is placed in a uniform magnetic field of stregth 0.001 T, then what torque will act on the coil, when the plane of the coil is inclined at 30^(@) with the magnetic field,

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Answer Text

Solution :
Number of turns 100 <br> Current `I=0.5A` <br> Radius of the circular coil, `r=5 cm =5xx10^(-2)m` <br> `:.` Area of the coil, `A= pi r^(2)=3.14xx(5xx10^(-2))m^(2)` <br> Magnetic field intensity, `B=0.001T`. <br> Now torque on the current carrying coill due to magnetic field `tau =nBIAsin theta` <br> When the coil makes an angle `30^(@)` with the field, then `theta=60^(@)` <br> So, `tau=nBIa sin 60^(@)` <br> `=10xx0.001xx0.5xx3.14xx25xx10^(-4) xx(sqrt(3))/(2)` <br>

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