If we did all things we are capable of, we would literally astound ourselves.

– Thomas A. Edison

Friday, September 26, 2008

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)2F22]

(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

Friday, September 12, 2008

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.