Question

A conducting square wire frame ABCD of side l
is pulled by horizontal force so that it moves
with constant velocity v. A uniform magnetic field of strength B is existing perpendicular to the plane of wire. The resistance per unit length of wire is lambda and negligible self inductance. If at t=0, frame is just at the boundary of magnetic field
- Potential difference across BC at time
is1. bvl/4 2. bvl 3. 3 bvl /4

Answer 1

Answer:

Explanation:

answer

zero

Answer 2

Concept

Magnetic Lorentz force: When a charge moves in a magnetic field, it experiences a magnetic force given by

F = q v × B,

where v is the velocity of the charge,

B is the magnetic field.

Given

• A square wireframe ABCD each side of which is l units.

• Resistance per unit length of the wire is lambda.

• Self inductance of ABCD is negligible.

• ABCD is moving towards the magnetic field region with constant velocity v.

• The magnetic field is perpendicular to the wireframe plane.

Find

Potential difference across side BC.

Solution

Let the magnetic field is set up in the z direction, the wireframe is moving in the x direction as shown in the figure.

Magnetic force acting on the electron

The force acting on the electrons in the side arms of BC will be in the y direction and so they do not move to produce current in that direction.

In the arm BC, the magnetic force on an electron is -e v × b = -evb $\mathbf{\hat i}$ × $\mathbf{\hat k}$ = -evb(-$\mathbf{\hat j}$) = evb $\mathbf{\hat j}$.

So, the electron moves in the positive y direction under the magnetic force of magnitude evb.

Work done per unit charge

This movement of electrons under magnetic force causes current in the wire and thus a potential difference across any two point on the wire.

The work done by the magnetic force in moving the electron from C to B is levb.

Instead of an electron moving from C to B we can equally assume a positive charge e to be moving from B to C. The work will be the same.

Now, the work done on unit positive charge in moving it from B to C is

                                W = levb/e = bvl.

Potential difference across BC

If the self inductance of the wireframe was not negligible, a significant back emf would have produced that opposes the current being established as the wireframe starts entering the magnetic field region.

This back emf would have prolonged the time for the establishment of steady current under the magnetic force. Due to this the potential difference across BC would not be equal to the work W.

Assuming the self inductance to be negligible, a steady current would be produced in no time and the work done W will be the potential difference between points B and C.

So, the potential difference across the side BC is bvl.

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