AIPMT 2008 Question on Magnetic force

Most of you might have noted that a magnetic dipole placed in a non uniform magnetic field generally experiences a net force and a torque. If the magnetic field is uniform, the net force will be zero; but there will be a torque (if the magnetic moment vector is not parallel or antiparallel to the magnetic field direction). The following question which appeared in AIPMT 2008 question paper is worth noting:

A closed loop PQRS carrying a current is placed in a uniform magnetic field. If the magnetic forces on segments PS, SR and RQ are F₁, F₂ and F₃ respectively and are in the same plane of the paper and along the directions shown, the force on the segment QP is

(1) √[(F₃ – F₁)² + F₂²]

(2) √[(F₃ – F₁)² – F₂²]

(3) F₃ – F₁ + F₂

(4) F₃ – F₁ – F₂

The net force on the loop in the horizontal direction is F₃ – F₁. Since the force F₂ is in the vertical direction, the reultant of the three forces F₁, F₂ and F₃ is √[(F₃ – F₁)² + F₂²]. Since the net force on the entire loop must be zero in the uniform magnetic field, the force on the segment QP must be the equilibrant of √[(F₃ – F₁)² + F₂²]. Therefore the force on the segment QP has magnitude √[(F₃ – F₁)² + F₂²] and direction opposite [Option (1)].

Now consider a question which is a little more difficult:

A proton of charge q and mass m proceeding along the positive X-direction with speed v encounters a uniform magnetic field of flux density B directed along the negative Z-direction. If the field is confined in the region between x = 0 and x = d and the proton emerges from the field along a direction making an angle of 45º with its initial velocity, the value of d must be

(a) (1/√2)(mv/qB)

(b) √2 mv/qB

(c) mv/qB

(d) 2 mv/qB

(e) mv/2qB

The situation is shown in the adjoining figure. C is the centre of the circular path of the proton in the magnetic field and AB is the radius R drawn from the point A from which the proton exits from the field. We have

R = mv/qB which you get by equating the centripetal force mv²/R to the magnetic force qvB.

Since d = AN = R sin45º where N is the foot of the perpendicular (drawn from A) to the Y-axis, we have

D = R/√2 = (1/√2)(mv/qB)

You will find many useful questions (with solution) on magnetic force at apphysicsresources