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Class 10 Class 12

Electromagnetic Induction

The current in a coil of self-inductance L = 2H is increasing according to the law i = 2 sin t².
Find the amount of energy spent during the period when the current changes from 0 to 2 ampere.

Given,
Self inductance of the coil, L = 2 H
Let the current flowing throught he coil be 2 ampere at time = t².

Then,

We have tocalculate, the amount of energy spent when current changes from 0 to 2 A.
Now,
Self induced emf is

L \frac{di}{dt} .

where, i is the instantaneous value of current.

If, dq is the amount of charge displaced in time dt then elementary work done

dW = L (\frac{di}{dt}) dq = L \frac{di}{dt} idt = Li di

Therefore,

W = \int_{0}^{τ} Li di = \int_{0}^{τ} L 2 \sin t^{2} d (2 \sin t^{2}) W = \int_{0}^{τ} 8 L \sin t^{2} \cos t^{2} t dt = 4 L \int_{0}^{τ} \sin 2 t^{2} t dt

Let,

θ = 2 t^{2} \Rightarrow dθ = 4 t dt \Rightarrow dt = \frac{dθ}{4 t}

∴

The integral becomes,

W

= 4 L \int \frac{\sin θ dθ}{4}

= L (- \cos θ) = - L \cos 2 t^{2}

W = - L {[\cos 2 t^{2}]}_{0}^{\sqrt{π} / 2} = 2 L = 2 \times 2 = 4 J .

That is, 4 Joule of nergy is spent when the current is raised to 2 Amperes.

693 Views

(a) State Lenz’s law. Give one example to illustrate this law. “The Lenz’s law is a consequence of the principle of conservation of energy.” Justify this statement.
(b) Deduce an expression for the mutual inductance of two long coaxial solenoids but having different radii and different number of turns.

(a) According to Lenz’s law, the direction of the induced current (caused by induced emf) is always such as to oppose the change causing it.

box enclose straight epsilon space equals space minus straight k dϕ over dt end enclose

where k is a positive constant. The negative sign expresses Lenz’s law. It means that the induced emf is such that, if the circuit is closed, the induced current opposes the change in flux.
(b) Figure shows a coil of N₂ turns and radius R₂ surrounding a long solenoid of length l₁radius R_l and number of turns N₁. To calculate M between them, let us assume a current i₁through the inner solenoid S₁There is no magnetic field outside the solenoid and the field inside has magnitude,

box enclose straight B space equals space straight mu subscript 0 open parentheses straight N subscript 1 over straight l subscript 1 close parentheses straight i subscript 1 end enclose

and is directed parallel to the solenoid’s axis. The magnetic flux Φ_B2 through the surrounding coil is, therefore,

straight ϕ subscript straight B 2 end subscript space equals space straight B open parentheses πR subscript 1 squared close parentheses space equals space fraction numerator straight mu subscript 0 straight N subscript 1 straight i subscript 1 over denominator straight l subscript 1 end fraction πR subscript 1 squared

Now,

straight M space equals space fraction numerator straight N subscript 2 straight ϕ subscript straight B 2 end subscript over denominator straight i subscript 1 end fraction space equals space open parentheses straight N subscript 2 over straight i subscript 1 close parentheses space open parentheses fraction numerator straight mu subscript 0 straight N subscript 1 straight i subscript 1 over denominator straight l subscript 1 end fraction close parentheses πR subscript 1 superscript 2 space equals space fraction numerator straight mu subscript 0 straight N subscript 1 straight N subscript 2 πR subscript 1 superscript 2 over denominator straight l subscript 1 end fraction comma space straight M space equals fraction numerator straight mu subscript 0 straight N subscript 1 straight N subscript 2 πR subscript 1 superscript 2 over denominator straight l subscript 1 end fraction

Notice that M is independent of the radius R₂ of the surrounding coil. This is because solenoid's magnetic field is confined to its interior.

1779 Views

What is induced emf? Write Faraday’s law of electromagnetic induction. Express it mathematically. A conducting rod of length L, with one end pivoted, is rotated with a uniform angular speed ‘ω’ in a vertical plane, normal to a uniform magnetic field ‘B’. Deduce an expression for the emf induced in this rod.

Whenever the magnetic flux linked with a closed circuit changes, an emf is set up across it which lasts only as long as the change in flux is taking place. This emf is called induced emf.

According to Faraday's law of electromagnetic induction, the magnitude of induced emf is equal to the rate of change of magnetic flux linked with the closed circuit (or coil).

Mathematically,
$E = - N \frac{{dϕ}_{B}}{dt}$

where,
N is the number of turns in the circuit and,
Φ_B is the magnetic flux linked with each turn.

Suppose the conducting rod completes one revolution in time T.
Then,
Change in flux = B x Area swept
= B x $π$ l²Therefore,

^{$Induced emf = \frac{Change in flux}{Time}$

$ε = \frac{B \times {πl}^{2}}{T}$}

But, $T = \frac{2 π}{ω}$

$∴$ $ε = \frac{B \times π l^{2}}{2 π / ω} = \frac{1}{2} {Bl}^{2} ω .$

1282 Views

A circular loop of radius 0.3 cm lies parallel to a much bigger circular loop of radius 20 cm. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is 15 cm. (a) What is the flux linking the bigger loop if a current of 2.0 A flows through the smaller loop? (b) Obtain the mutual inductance of the two loops.

Given here,
Radius of bigger loop, r₁= 20 cm
Radius of small circular loop, r₂ = 0.3 cm

Distance betwen the centers of smaller and bigger loop, x = 15 cm

We know from the considerations of symmetry that M₁₂ = M₂₁.

Direct calculation of flux linking the bigger loop due to the field by the smaller loop will be difficult to handle. Instead, let us calculate the flux through the smaller loop due to a current in the bigger loop. The smaller loop is so small in area that one can take the simple formula for field B on the axis of the bigger loop and multiply B by the small area of the loop to calculate flux without much error.

Let 1 refer to the bigger loop and 2 the smaller loop.
Field B₂ at loop 2 due to crrent I₁ in loop 1 is,

B_{21} = \frac{μ_{0} I_{1} r_{1}^{2}}{2 {(x^{2} + r_{1}^{2})}^{3 / 2}}

Here x is distance between the centres.

Thus,
Flux on loop 2 due to current in loop 1 is,

ϕ_{21} = B_{21} \times A_{2}

ϕ_{21} = B_{2} {πr}_{2}^{2} = \frac{π μ_{0} r_{1}^{2} r_{2}^{2}}{2 {(x^{2} + r_{1}^{2})}^{3 / 2}} I_{1}

But,

ϕ_{21} = M_{21} I_{1}

∴

M_{21} = \frac{{πμ}_{0} r_{1}^{2} r_{2}^{2}}{2 {(x^{2} + r_{1}^{2})}^{3 / 2}} = M_{12}

and,

ϕ_{12} = M_{12} I_{2} = \frac{{πμ}_{0} r_{1}^{2} r_{2}^{2}}{2 {(x^{2} + r_{1}^{2})}^{3 / 2}} I_{2}

Therefore,

M_{12} = M_{21} = \frac{{πμ}_{0} r_{1}^{2} r_{2}^{2}}{2 {(x^{2} + r_{1}^{2})}^{3 / 2}}

Numerical:

Using the given data
b) Mutual inducatance, M₁₂ = M₂₁ =

\frac{π μ_{0} r_{1}^{2} r_{2}^{2}}{2 {(x^{2} + r_{1}^{2})}^{3 / 2}}

\frac{π \times 4 π \times 10^{- 7} \times (20 \times 10^{- 2})^{2} \times (0.3 \times 10^{- 2})^{2}}{2 \times {[(15 \times 10^{- 2})^{2} + (20 \times 10^{- 2})^{2}]}^{\frac{3}{2}}}

= 4.55 x 10^–11 H

a) Flux linking with the bigger loop when I₁ is 2.0 A is given by,

ϕ_{21} = M_{21} I_{1}

= 4.55 x 10^–11 x 2
= 9.1 x 10^–11 Wb

285 Views

A 0.5 m long metal rod PQ completes the circuit as shown in the figure. The area of the circuit is perpendicular to the magnetic field of flux density 0.15 T. If the resistance of the total circuit is 3 Ω, calculate the force needed to move the rod in the direction as indicated with a constant speed of 2 m s^–1.

Consider coil PQRS with its arm PQ movable as shown in the figure. A magnetic field is applied normal to the surface of the coil. The area of the coil, ΔS = l x x
Φ = 8 Δ S = Blx
The rate of change of magnetic flux linked with the coil is given by

   $dϕ over dt space equals space fraction numerator straight d open parentheses Blx close parentheses over denominator dt end fraction space space space space space space space space equals space Bl dx over dt equals Blv$
If e is the induced emf produced, then
   $straight e space equals negative dϕ over dt space equals space Blv$
$open square brackets because space straight x space is space decreasing comma space fraction numerator straight d left parenthesis straight x right parenthesis over denominator dt end fraction space is space minus space Ve close square brackets$
Let R be the resistance of movable arm PQ of the rectangular conductor. Taking the resistance of other arms as negligibly small, the current in the loop is given by,
I = $straight epsilon over straight R space equals Blv over straight R$ ...(I)
The arm PQ moves with the speed v. The power required to move it is given by,
   $straight F space equals space BIL$
$therefore$    $straight F space equals space fraction numerator Bl squared straight v over denominator straight R end fraction$ (From I)
$rightwards double arrow$    $straight F space equals space fraction numerator straight B squared straight l squared straight v over denominator straight R end fraction space equals space fraction numerator left parenthesis 0.15 right parenthesis space left parenthesis 0.5 right parenthesis squared space 2 space over denominator 3 end fraction$
$equals space 3.75 space cross times space 10 to the power of negative 3 end exponent straight N.$

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Electromagnetic Induction