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

Two Kerala Engineering & Medical Entrance 2008 Questions on Resistive Networks

Questions involving resistive networks can be solved are often found in degree entrance examinations. Here is a reasonably good question which appeared in KEAM 2008 (Engineering) question paper:

In the circuit shown, if the resistance 5 Ω develops a heat of 42 J per second, the heat developed in 2 Ω must be about (in J s^–1)

(a) 25

(b) 20

(c) 30

(d) 35

(e) 40

The heat developed per second (power dissipation) in a resistance R is V²/R where V is the voltage across the resistance

[This expression has the forms I²R and VI where I is the current. Even though you can use these forms also to solve the problem, the form V²/R will be more convenient here]

Therefore we have

V²/5 = 42 watt

The power dissipated in the parallel branch containing the resistors 6 Ω and 9 Ω is V²/(6+9) = V²/15 and this must be 14 watt (since the denominator is 3 times)

The total power dissipated in the two parallel branches is thus 42+14 = 56 watt and the effective resistance of the two parallel branches is 5×15/(5+15) = 75/20 Ω.

If I is the current passing through the 2 Ω resistance, we have

I²×2 = W where W is the power dissipated in 2 Ω.

Since the same current passes through the effective resistance (75/20 Ω) of the parallel branches, we have

I²×75/20 = 56

From the above equations (on dividing),

40/75 = W/56 so that W = 40×56/75 = 30 watt nearly

The following MCQ which appeared in KEAM 2008 (Medical) question paper is simple:

In the Wheatstone’s network shown in the figure, the current I in the circuit is

(a) 1 A

(b) 2A

(c) 0.25 A

(d) 0.5 A

(e) 0.33 A

The bridge is balanced (since 2 Ω/4 Ω = 4 Ω/8 Ω). Therefore, the points B and D are equipotential points and there is no current through the diagonal 5 Ω resistor. The circuit thus reduces to 6 Ω and 12 Ω in parallel with the 2 V battery. The parallel combined value of 6 Ω and 12 Ω is 6×12/(6+12) = 4 Ω.

The current, I =2 V/4 Ω = 0.5 A.