A 100 turn closely wound circular coil of radius 10 cm carries a current of 3.2 A. (a) What is the field at the centre of the coil? (b) What is the magnetic moment of this coil? The coil is placed in a vertical plane and is free to rotate about a horizontal axis which coincides with its diameter. A uniform magnetic field of 2T in the horizontal direction exists such that initially the axis of the coil is in the direction of the field. The coil rotates through an angle of `90^@` under the influence of the magnetic field. (c) What are the magnitudes of the torques on the coil in the initial and final position? (d) What is the angular speed acquired by the coil when it has rotated by `90^@`? The moment of inertia of the coil is` 0.1 kg m^2`.

Explanation

(a) Magnetic field produced at the centre of current carrying circular ring having N no. of closely wound turns is,
`B=(mu_(0)NI)/(2R)`
`thereforeB=((4pixx10^(-7))(100)(3.2))/((2)(0.1))`
`thereforeB=2.0096xx10^(-3)T`
(b) Magnetic moment of given coil,
`vecm=NIvecA`
`thereforem=NIA`
`=NI(piR^(2))`
`=(100)(3.2)(3.14)(0.1)^(2)`
`thereforem=10.048Am^(2)`
( c) (i) Initial position :
Here angle `theta` between

`vecAandvecB" is "theta=0^(@)`.
Hence, torque exerted on the coil will be,
`tau=mBsintheta`
= `mBsin0^(@)`
`thereforetau=0" "(becausesin0^(@)=0)`
(ii) Final position :

Here angle `theta` between `vecAandvecB" is " theta=90^(@)`
Hence, torque exerted on the coil will be,
`tau=mBsintheta`
= `(10.048)(2.0096xx10^(-3))sin90^(@)`
`thereforetau=20.19Nm`
(d) We have `tau=mBsintheta`
Amount of work done by above torque during extremely small angular displacement `d theta` will be,
`dW=taud theta`
`thereforedW=mBsinthetad theta`
`thereforeintdW=mBintsinthetad theta`
`thereforeW=mB{-costheta}_(theta_(1))^(theta_(2))`
`thereforeW=mB(-costheta_(2)+costheta_(1))`
`thereforeW=mB(costheta_(1)-costheta_(2))`
`thereforeDeltaK=mB(costheta_(1)-costheta_(2))`
(`because` According to work energy theorem)
`therefore1/2Iomega_(2)^(2)-1/2Iomega_(1)^(2)=mB(costheta_(1)-costheta_(2))`
`therefore1/2Iomega_(2)^(2)-0=mB(cos0^(@)-cos90^(@))`
`(because"Here",omega_(1)=0" "theta_(1)=0^(@),theta_(2)=90^(@))`
`therefore1/2Iomega_(2)^(2)=mB`
`thereforeomega_(2)^(2)=(2mB)/I`
`thereforeomega_(2)=sqrt((2mB)/I)`
`thereforeomega_(2)=sqrt((2xx10.048xx2)/0.1)" "(because"Here",B=2T)`
`thereforeomega_(2)=20.048rad//s`