(a)Derive the expression for the torque on a rectangular current carrying loop suspended in a uniform magnetic field. (b)A proton and a deuteron having equal momenta enter in a a region of a uniform magnetic field at right angle to the direction of a the field.Depict their trajectories in the field. OR (a)A small compass needle of magnetic moment m is free to turn about an axis perpendicular to the direction of unifrom magnetic field B.The moment of inertia of the needle about the axis is I.The needle is slightly disturbed form its stable position and then released.Prove that it executes simple harmonic motion.Hence deduce the expression for its time period. (b) A compass needle, free to turn in a vertical plane orients inself with its axis vertical at a certain place on the earth.Find out the values of (i)horizontal component of earth's magnetic field and (ii) angle of dip at the place.

(a)Derive the expression for the torque on a rectangular current carry

1.	(a)Derive the expression for the torque on a rectangular current carrying loop suspended in a uniform magnetic field. (b)A proton and a deuteron having equal momenta enter in a a region of a uniform magnetic field at right angle to the direction of a the field.Depict their trajectories in the field. OR (a)A small compass needle of magnetic moment m is free to turn about an axis perpendicular to the direction of unifrom magnetic field B.The moment of inertia of the needle about the axis is I.The needle is slightly disturbed form its stable position and then released.Prove that it executes simple harmonic motion.Hence deduce the expression for its time period. (b) A compass needle, free to turn in a vertical plane orients inself with its axis vertical at a certain place on the earth.Find out the values of (i)horizontal component of earth's magnetic field and (ii) angle of dip at the place.
Answer» Solution :Where `theta=` angle between velocity of particle and magnetic FIELD `=90^(@)` So,Lorentz force, `F=BqV` Thus the particles will move in circular path. `Bqupsilon=(mV^(2))/r rArrr=(mV)/(Bq)` Let `m_(p)=` mass of proton `m_(d)=` mass of deuteron, `V_(p)=` velocity of proton and `V_(d)=`velocity of deuteron.The CHARGE of proton and deuteron are equal. Given that `m_(p)V_(p)=m_(d)V_(d)` `r_(p)=(m_(p)upsilon_(p))/(Bq)`...(i) `r_(d)=(m_(d)upsilon_(d))/(Bq)`...(ii) As (i) and (ii) are equal so `r_(p)=r_(d)=r` Thus, the trajectory of both the particles will be same. OR (b)(i)As Horizontal component of earth's magnetic field `B_(H)=Bcos delta` Putting `=delta=90^(@) therefore B_(H)=0` (ii)For a compass NEEDLE align vertical at a certain place, angle of DIP , `delta=90^(@)`

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