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CBSE Class 12-science Answered

(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 region of uniform magnetic field at right angle to the direction of 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 uniform magnetic field 'B'. The moment of inertia of the needle about the axis is 'I'. The needle is slightly disturbed from 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 a turn in a vertical plane orients itself 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.
Asked by Topperlearning User | 01 Jun, 2015, 02:10: PM
answered-by-expert Expert Answer

(b)

A proton is a positively charged particle. So it will execute a circular trajectory.

Deuteron is uncharged particle. So it will not be deflected in the magnetic field.

 

OR,

So

(b)

Answered by | 01 Jun, 2015, 04:10: PM
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