Two long parallel horizontal rails, distance d apart and each having a resistance A per unit length, are joined at one end by a resistance R. A perfectly conducting rod MN of mass m is free to slide along the rails without friction (see Fig.). There is a uniform magnetic field of induction B normal to the plane of the paper and directed into the paper. A variable force F is applied to the rod MN such that as the rod moves, constant current flows through R.
(i) Find the velocity of the rod and the applied force F as function of the distance x of the rod from R.
(ii) What fraction of the work done per second by F is converted into heat?

Solution

Let, the distance from R to MN be x.

Then the area of the loop between MN and R is xd and,
Magnetic flux linked with the loop is B x d.

As the rod moves, the emf induced in the loop is given by

|e| = \frac{d}{dt} (B x d) = Bd \frac{dx}{dt} = B v d

where v is the velocity of MN.

The total resistance of the loop between R and MN is R + 2λx.

The current in the loop is given by

i = \frac{|e|}{R + 2 λx} = \frac{Bvd}{R + 2 λx}

(i) Force acting on the rod,

F = iBd = \frac{B^{2} d^{2}}{R + 2 λx} v

∴

m \frac{dv}{dt} = \frac{B^{2} d^{2}}{R + 2 λx} . \frac{dx}{dt}

dv = \frac{B^{2} d^{2}}{m} . \frac{dx}{R + 2 λx}

On integrating both sides, we get

ν = \frac{B^{2} d^{2}}{2 λm} In (\frac{R + 2 λx}{R})

and,

force = \frac{B^{2} d^{2}}{R + 2 λx} . \frac{B^{2} d^{2}}{2 λm} In (\frac{R + 2 λx}{R})

(ii) Work done per second = Fv

Heat produced per second =

i^{2} (R + 2 λx)

= {(\frac{Bvd}{R + 2 λx})}^{2} (R + 2 λx)

= (\frac{B^{2} d^{2} ν}{R + 2 λx}) . ν = F . ν

Thus, the ratio of heat produced to work done is 1.
The entire work done by F per second is converted into heat.

Some More Questions From Electric Charges and Fields Chapter

a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?
b) Explain why two field lines never cross each other at any point?

Figure shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?

For Daily Free Study Material Join wiredfaculty

Electric Charges And Fields

Some More Questions From Electric Charges and Fields Chapter

a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?
b) Explain why two field lines never cross each other at any point?

Figure shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction

For Daily Free Study Material Join wiredfaculty

Electric Charges And Fields

Some More Questions From Electric Charges and Fields Chapter

a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?b) Explain why two field lines never cross each other at any point?

Figure shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?
b) Explain why two field lines never cross each other at any point?