Electromagnetic Induction

Q: A closed circuit consists of a source of constant emf E and a choke coil of inductance L connected in series. The active resistance of the whole circuit is equal to R . At the moment t = 0 the choke coil inductance was decreased abruptly \eta times. Find the current in the circuit as a function of time t .Instruction. During a stepwise change of inductance the total magnetic flux (flux linkage) remains constant.

We write the equation of the circuit as, Ri = \dfrac {L}{\eta} \dfrac {di}{dt} = \xi ,for t\geq 0 . The current at t = 0 just after inductance is changed, is i = \eta \dfrac {\xi}{R} , so that the flux through the inductance is unchanged.We look for a solution in the form i = A + Be^{-t/C} Substituting C = \dfrac {L}{\eta R}, B = \eta - 1, A = \dfrac {\xi}{R} Thus, i = \dfrac {\xi}{R} (1 + (\eta - 1)e^{-\eta Rt/ L}) .

Q: Find the mutual inductance between the circular coil of radius a^{'} and the loop of radius a as shown in figure when the slider on the rheostat is moved. The total resistance of rheostat is R and the distance between the center of two coils is x.

Refer image,The magnetic field at centre of coil 2, due to coil 1 is given by: B=\dfrac{\mu_0\,ia^2}{2(a^2+x^2)^{3/2}} The fluxed linked with coil 2 is given by \Phi=BA' =\dfrac{\mu_0\,ia^2}{2(a^2+x^2)^{3/2}}\pi a^{1^2} Now, let y be the distance of the sliding contact from its left end.Given: v=\dfrac{dy}{dt} Total Resistance of rheostat = RWhen the distance of the sliding contact from the left end is y, the resistance of rheostat is given by r'=\dfrac{R}{L}y The current in the coil is the function of distance y travelled by the sliding contact of the rheostat. It is given by i=\dfrac{E}{\left(\dfrac{Ry}{L}+r\right)} The magnitude of emf induced can be calculated as e=\dfrac{d\Phi}{dt}=\dfrac{\mu_0a^2a^{1^2}\pi}{2(a^2+x^2)^{3/2}}\dfrac{di}{dt} e=\dfrac{\mu_0\,\pi a^2a^{1^2}}{2(a^2+x^2)^{3/2}}\dfrac{d}{dt}\dfrac{E}{\left(\dfrac{R}{L}y+r\right)} e=\dfrac{\mu_0\,a^2a^{1^2}}{2(a^2+x^2)^{3/2}}E\left[\dfrac{\left(-\dfrac{R}{L}v\right)}{\left(\dfrac{R}{L}y+r\right)^2}\right] emf induced, e=\dfrac{\mu_0\pi \,a^2a^{1^2}}{2(a^2+x^2)^{3/2}}E\left[\dfrac{-\dfrac{R}{L}v}{\left(\dfrac{Ry}{L}y\right)^2}\right] Now, emf induced in coil can be also given as \dfrac{di}{dt}=\left[\dfrac{E\left(-\dfrac{R}{L}v\right)}{\left(\dfrac{Ry}{L}y\right)^2}\right] e=\dfrac{Mdi}{dt} \dfrac{di}{dt}=\left[\dfrac{E\left(-\dfrac{Rv}{L}\right)}{\left(\dfrac{Ry}{L}+r\right)^2}\right] \therefore M=\dfrac{e}{\dfrac{di}{dt}}=\dfrac{\mu_0\pi a^2a^{1^2}}{2(a^2+x^2)^{3/2}}

Q: An ac generator has emf \xi=\xi _{m}\sin \omega_{d}t , with \xi _{m}=25.0V and \omega _{d}=377rad/s . It is connected to a 12.7 H inductor. (a) What is the maximum value of the current? (b) When the current is a maximum, what is the emf of the generator? (c) When the emf of the generator is -12.5 V and increasing in magnitude, what is the current?

(a) The circuit consists of one generator across one inductor; therefore, \varepsilon _{m}=V_{L} . The current amplitude is I_m=\dfrac{\varepsilon _{m}}{X_{L}}=\dfrac{\varepsilon _{m}}{\omega _{d}L}=\dfrac{25.0V}{\left ( 377rad/s \right )\left ( 12.7H \right )}=5.22 \times 10^{-3}A .(b) I = I_m sin(\omega _d t - \phi) --(i)Current will be maximum when sin(\omega _d t - \phi) = 1 or, \omega _d t - \phi = \dfrac{\pi}{2} At t = 0 potential difference equals zero as sin(\omega _d t) = 0 \phi = -\dfrac{\pi}{2} This is the phase angle between current and potential difference.Now, I = I_m sin(\omega_d t + \dfrac{\pi}{2}) or, I = I_m(cos \omega _d t) So, maximum current occurs when cos(\omega _d t) = 1 That gives t = 0 At this instant potential difference is zero: \varepsilon = 0 (c) Consider equation \xi =\xi _{m}\sin \omega _{d}t with \varepsilon =-\dfrac{\varepsilon _{m}}{2} . In order to satisfy this equation, we require \sin \left ( \omega _{d}t \right )=-1/2 . Now we note that the problem states that ε is increasing in magnitude, which (since it is already negative) means that it is becoming more negative. Thus, differentiating equation with respect to time (and demanding the result be negative) we must also require \cos \left ( \omega _{d}t \right )< 0 . These conditions imply that \omega t must equal \left ( 2n\pi -5\pi /6 \right ) [ n = integer]. Consequently, equation i=I\sin \left ( \omega _{d}t+\phi\right) yields (for all values of n ) i=I\sin \left ( 2n\pi -\frac{5\pi}{6}-\frac{\pi}{2} \right )=\left ( 5.22 \times 10^{-3}A \right )\left ( \frac{\sqrt{3}}{2} \right )=4.51 \times 10^{-3}A .

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Current induced in a circuit due to magnetic flux.
Find the current (as a function of time t) in the circuit consisting of a source of constant emf E and a choke coil of inductance L connected in series.

A closed circuit consists of a source of constant emf

E

and a choke coil of inductance

L

connected in series. The active resistance of the whole circuit is equal to

R

. At the moment

t = 0

the choke coil inductance was decreased abruptly

η

times. Find the current in the circuit as a function of time

t

.
Instruction. During a stepwise change of inductance the total magnetic flux (flux linkage) remains constant.

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The magnetic flux through a stationary loop with resistance R varies during the interval of time T as $ϕ = a t (T - t)$ ./ The heat generated during this time neglecting the inductance of the loop will be :

\frac{a ^{2} T ^{3}}{3 R}

\frac{a ^{2} T ^{2}}{3 R}

\frac{a ^{2} T}{3 R}

\frac{a ^{3} T ^{3}}{3 R}

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Problem on faraday law and Lenz's law.
Faraday's Laws of Electromagnetic Induction consists of two laws. The first law describes the induction of EMF in a conductor and the second law quantifies the EMF produced in the conductor. According to Lenz's Law, if an induced current flows in a coil due to electromagnetic induction, its direction is always such that it will oppose the change which produced it.

Three resistance of magnitude

R

each are connected in the form of an equilateral triangle of side

a

. The combination is placed in a magnetic field

B = B_{0} e^{- λ t}

perpendicular to the plane. The induced current in the circuit is given by:

(\frac{a ^{2} λ}{2 3 R} B_{0}) e^{- λ t}

(\frac{a ^{2} λ}{4 ( 3 ) R} B_{0}) e^{- λ t}

(\frac{a ^{2} B _{0}}{λ 4 3 R}) e^{- λ t}

(\frac{a ^{2} B _{0} R}{λ 4 3 R}) e^{- λ t}

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Steady state Current.
A system (e.g. circuit) is in the steady state when the current at each point in the circuit is constant (does not change with time). In many practical circuits, the steady state is achieved in a short time.

A solenoid of inductance

L

and resistance

r

is connected in parallel to a resistance

R

and a battery of emf

E

. Initially if the switch is closed for a long time and at

t = 0

, then the :

current through solenoid at any time

t

, after opening the switch is

\frac{E}{r} e^{- (R + r) t / L}

induced emf across solenoid at time

t = 0

\frac{E ( R + r )}{r}

amount of heat generated in solenoid is

\frac{E ^{2} L}{2 r ( r + R )}

potential difference across solenoid at

t = 0

E

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Mutual inductance
Mutual Inductance is the interaction of one coils magnetic field on another coil as it induces a voltage in the adjacent coil.

Find the mutual inductance between the circular coil of radius

a^{^{'}}

and the loop of radius

a

as shown in figure when the slider on the rheostat is moved. The total resistance of rheostat is R and the distance between the center of two coils is x.

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Energy stored in an inductor due to magnetic field.
The energy stored in a magnetic field is equal to the work needed to produce a current through the inductor having self inductance (L).

A solenoid of

10

henry inductance and

2

ohm resistance is connected to a

10

volt battery. In how much time the magnetic energy will be reaches to

\frac{1}{4}

th of the maximum value?

3.5 sec

2.5 sec

5.5 sec

7.5 sec

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EMF And Current In AC Generator
AC generators work on the principle of Faraday's law of electromagnetic induction which states that electromotive force -EMF or voltage - is generated in a current-carrying conductor that cuts a uniform magnetic field. Let's solve a problem to understand it properly.

An ac generator has emf

ξ = ξ_{m} sin ω_{d} t

, with

ξ_{m} = 25.0 V

and

ω_{d} = 377 r a d / s

. It is connected to a

12.7 H

inductor. (a) What is the maximum value of the current? (b) When the current is a maximum, what is the emf of the generator? (c) When the emf of the generator is

- 12.5 V

and increasing in magnitude, what is the current?

View solution