Question

In the following diagram, the arm PQ of the rectangular conductor is moved from x=0; outwards. The uniform magnetic field is perpendicular to the plane and extends from x=0 to x=b and is zero for x>b.
Only the arm PQ possesses substantial resistance 'r'. Consider the situation when the arm PQ is pulled outwards from x=0 to x=b, and is then moved back to x=0 with constant speed 'v'. Obtain expressions for the
(1) electric flux
(2) the induced emf
(3) the force necessary to pull the arm and
(4) the power dissipated qs joule heat.
sketch the variation of these quantities with distance.​

Answer 1

Answer:

1) Flux linked with the loop is given as

$d\phi = B.(L vdt)$

2) Induced EMF is given as

$\frac{d\phi}{dt} = v B L$

3) Force necessary to pull the arm is given as

$F = \frac{vB^2L^2}{r}$

4) Power dissipated as heat across the rod is given as

$P = \frac{v^2B^2L^2}{r}$

Explanation:

As we know by Faraday's law that rate of change in flux will induce EMF

so we will have

1) Magnetic flux is given as

$\phi = B.A$

$d\phi = B.(L vdt)$

2) Induced EMF is given as

now rate of change in flux is given as

$\frac{d\phi}{dt} = v B L$

here v = speed of the rod

L = length of the rod

B = constant magnetic field

3) Force necessary to pull the arm is given as

$F = i L B$

$F = \frac{vBL}{r} LB$

$F = \frac{vB^2L^2}{r}$

4) Power dissipated as heat across the rod is given as

$P = i^2 R$

$P = (\frac{vBL}{r})^2 r$

$P = \frac{v^2B^2L^2}{r}$

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Topic : Faraday's law of induced EMF

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Answer 2

Electric flux, induced emf, force and heat loss

Explanation:

For forward motion from x = 0 to x = 2b:

The flux ϕB linked with circuit SPQR is  ϕB = Blx

0 ≤ x < b

Blb

b ≤ x < 2b

The induced emf e = - dϕB/dt = - Blv ( 0 ≤ x < b)  = 0

b ≤ x < 2b

When induced emf is non-zero, the current İ in the magnitude.

I = e/r = Blv/r

The force required to keep arm PQ in constant motion is F =IlB.

Its direction is to the left.

In magnitude F = I/B = B2l2v/ r  

0 ≤ x < b

 = 0

b ≤ x < 2b  

The Joule heating loss is  PJ = I2r = B2l2v2/r  

0 ≤ x < b

= 0  

b ≤ x < 2b

One obtains similar expressions for the inward motion from x = 2b to x = 0

Please refer to the attached picture for the diagram.