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Class 10 Class 12

Electromagnetic Induction

Define magnetic flux. Give its SI unit.

The total number of magnetic lines of force crossing any surface in the magnetic field is termed as magnetic flux.

∴ Φ = BA cos θ

SI unit of magnetic flux is weber.

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A cylindrical bar magnet is kept along the axis of a circular coil. Will there be a current induced in the coil if the magnet is rotated about its axis? Give reasons.

No, current will not be induced in the coil if the coil is rotated about it's axis.

Formula for flux is $ϕ = NBA = constant$ .

$∴$ $e = \frac{dϕ}{dt} = 0 i = 0$

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How does the self-inductance of a coil change when an iron rod is introduced in the coil?

Since L ∝ μ_r , the self-inductance shall increase when an iron rod is introduced in the coil.

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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?

Let the distance from R to MN be x. Then the area of the loop between MN and R is xd and the magnetic flux linked with the loop is B x d. As the rod moves, the emf induced in the loop is given by

open vertical bar straight epsilon close vertical bar space equals space straight d over dt open parentheses straight B space straight x space straight d close parentheses space equals space Bd space dx over dt space equals space straight B space straight v space straight 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

straight i space equals space fraction numerator open vertical bar straight e close vertical bar over denominator straight R plus 2 λx end fraction space equals space fraction numerator Bvd over denominator straight R plus 2 λx end fraction

(i) Force acting on the rod,

straight F space equals space iBd space equals space fraction numerator straight B squared straight d squared over denominator straight R plus 2 λx end fraction straight v

therefore

straight m space dv over dt space equals space fraction numerator straight B squared straight d squared over denominator straight R plus 2 λx end fraction. dx over dt

or,

dv space equals space fraction numerator straight B squared straight d squared over denominator straight m end fraction. fraction numerator dx over denominator straight R plus 2 λx end fraction

On integrating both sides, we get

straight nu space equals space fraction numerator straight B squared straight d squared over denominator 2 λm end fraction In open parentheses fraction numerator straight R plus 2 λx over denominator straight R end fraction close parentheses

and

force space equals space fraction numerator straight B squared straight d squared over denominator straight R plus 2 λx end fraction. fraction numerator straight B squared straight d squared over denominator 2 λm end fraction In open parentheses fraction numerator straight R plus 2 λx over denominator straight R end fraction close parentheses

(ii) Work done per second = Fv
Heat produced per second =

straight i squared left parenthesis straight R plus 2 λx right parenthesis

Let the distance from R to MN be x. Then the area of the loop between

equals space open parentheses fraction numerator Bvd over denominator straight R plus 2 λx end fraction close parentheses squared space space left parenthesis straight R plus 2 λx right parenthesis

equals space open parentheses fraction numerator straight B squared straight d squared straight nu over denominator straight R plus 2 λx end fraction close parentheses. straight nu equals space straight F. straight nu

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

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Two different coils have self-inductances L₁ = 8 mH and L₂ = 2 mH. At a certain instant, the current in the two coils is increasing at the same constant rate and the power supplied to the two coil is the same. Find the ratio of (a) induced voltage (b) current and (c) energy stored in the two coils at that instant?

Given, two different coils.
L₁ = 8 mH; L₂ = 2 mH

a) Now, using the formula,
$e = L \frac{dI}{dt} \Rightarrow \frac{e_{1}}{e_{2}} = \frac{L_{1}}{L_{2}} = \frac{8}{2} = 4$

Hence, ratio of induced volatge is 4:1 .

b) Since, $P = eI = constant$

Therefore, $\frac{{dI}_{1}}{dt} = \frac{{dI}_{2}}{dt}$

$P_{1} = P_{2} = P$

$e_{1} I_{1} = e_{2} I_{2}$

$∴$ $\frac{I_{1}}{I_{2}} = \frac{e_{2}}{e_{1}} = \frac{1}{4}$
Ratio of current is 1:4 .

c) Energy is given by, $U = \frac{1}{2} {LI}^{2}$

$∴$ $\frac{U_{1}}{U_{2}} = \frac{\frac{1}{2} L_{1}}{\frac{1}{2} L_{2}} {(\frac{I_{1}}{I_{2}})}^{2} = \frac{8}{2} {(\frac{1}{4})}^{2} = \frac{1}{4} .$
Thus ratio of energy is 1 :4 .

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Electromagnetic Induction

How does the self-inductance of a coil change when an iron rod is introduced in the coil?