Electromagnetic Induction

Q: Derive an expression for the total magnetic energy stored in two coils with inductance 'L_1' and 'L_2' and mutual inductance 'M' ,when the currents in the coils are 'I_1' and 'I_2' respectively.

When the currents are increasing in the circuit, we have e.m.f.s E_1=-L_1\dfrac{dI_1}{dt} \pm M \dfrac{dI_2}{dt},E_2=-L_2\dfrac{dI_2}{dt} \pm M\dfrac{dI_1}{dt} where \pm sign appears consistently in both equations and depends on the geometry of the coils and the presence of current.Work done in pushing charges dq_1 and dq_2 through each circuit, respectively, is dW=-E_1dq_1-E_2 dq_2 =L_1\dfrac{dI_1}{dt}dq_1 \mp M \dfrac{dI_2}{dt}dq_1+L_2\dfrac{dI_2}{dt}dq_2 \mp M\dfrac{dI_1}{dt}dq_2 I_1=\dfrac{dq_1}{dt},dq_1=I_1 dt, I_2=\dfrac{dq_2}{dt},dq_2=I_2 dt, dW=L_1I_1dI_1+L_2I_2dI_2\mp (MI_1dI_2+MI_2dI_1) =L_1I_1dI_1+L_2I_2dI_2 \mp Md(I_1I_2) On integrating the above expression from 0 to final current i.e I_2 ,we have U=\int dW=L_1 \int^{I_1}_0I_1dI_1+L_2\int^{I_2}_0 dI_2 \mp M \int^{I_1I_2}_0 d(I_1I_2) =\dfrac{1}{2}L_1I^2_1 +\dfrac{1}{2}L_2I^2_2 \mp MI_1I_2

Q: A metallic rod of length 'I' is rotated with a frequency v with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius r, about an axis passing through the centre and perpendicular to the plane of the ring. A constant uniform magnetic field B parallel to the axis is present everywhere. Using Lorentz force, explain how emf is induced between the centre and the metallic ring and hence obtained the expression for it.

Suppose the length of the rod is greater than the radius of the circle and rod rotates anticlockwise and suppose the direction of electrons in the rod at any instant be along +y direction .Suppose the direction of the magnetic field is along +z direction.Then, using Lorentz law, we get the following \vec{F}=-e(\vec{v}\times \vec{B}) F=-e(v\hat{j}\times B\hat{k}) F¯=−e(v¯×B¯) \vec{F}=-evB\hat{i} Thus, the direction of force on the electrons is along -x axis. So, the electrons will move towards the centre i.e the fixed end of the rod. This movement of electron will effect in current and thus it will generate an emf in the rod between the fixed end and the point touching the ring. Let \theta be the angle between the rod inside the circle =\dfrac{1}{2}\pi r^2\theta =12πr2θInduced emf =B\times \dfrac{d}{dt}(\dfrac{1}{2}\pi r^2\theta)=\dfrac{1}{2}\pi r^2B\dfrac{d\theta}{dt}=\dfrac{1}{2}\pi r^2B\omega

Q: Figure shows a rectangular conductor PQRS in which the Conductor PQ is free to move in a uniform magnetic field B perpendicular to the plane of the paper. The filed extends from x=0 to x=b and is zero for x>b . Assume that only the arm PQ possesses resistance r. when the arm PQ is pulled outward from x=0 to x=2b and is then moved back to x=0 with constant speed v, Determine the expression for the flux and induced emf. Sketch the variations of these quantities with distance 0\le x\le 2b .

Let us consider forward motion from x = 0 to x = 2b , the flux \phi_B linked with the circuit SQPR is \phi_B = Blx 0<x<b = Blb b<x<2b The induced emf is E = \cfrac{-d\phi}{dt} \therefore E = BlV 0<x<b =0 b<x<2b

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Tricky questions

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- Some questions are often misinterpreted by the students. Let's explore them!

We often make a mistake in finding the direction of induced emf where the magnetic field is varying in space. For example, the magnetic field due to a current carrying wire increases as we move closer to it. So finding the induced emf in a loop can be difficult. But its direction is always consistent with the Lenz's law.

Conceptual Questions(10)
A loop of wire is moving near a long, straight wire carrying a constant current

I

as shown in Figure.
(a) Determine the direction of the induced current in the loop as it moves away from the wire.
(b) What would be the direction of the induced current in the loop if it were moving toward the wire?

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While writing the total magnetic energy stored in two coils with inductance 'L1' and 'L2' and mutual inductance 'M', we often tend to forget to involve the effect of mutual inductance. The energy stored is not just the sum of $\frac{L _{1} I _{1}^{2}}{2}$ and $\frac{L _{2} I _{2}^{2}}{2}$ .

Derive an expression for the total magnetic energy stored in two coils with inductance

^{'} L_{1}^{'}

and

^{'} L_{2}^{'}

and mutual inductance

^{'} M^{'}

,when the currents in the coils are

^{'} I_{1}^{'}

and

^{'} I_{2}^{'}

respectively.

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In problems based on rotational emf in a rod rotating with its one end fixed, it is often difficult to visualise the direction of the emf induced in the rod. We might make a mistake in denoting which end is at a higher potential. To find it correctly, we consider the elementary piece of the rod and find the direction of induced emf in it while considering its instantaneous linear velocity. We apply Lorentz force for that elementary part. The direction of induced emf will be same in all the elements and will get added up.

A metallic rod of length 'I' is rotated with a frequency v with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius r, about an axis passing through the centre and perpendicular to the plane of the ring. A constant uniform magnetic field B parallel to the axis is present everywhere. Using Lorentz force, explain how emf is induced between the centre and the metallic ring and hence obtained the expression for it.

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For plotting induced emf in a circuit whose area or flux is changing at a constant rate, we might make a mistake. It is to be noted that induced emf is ALWAYS associated with a changing flux. Zero flux means the induced emf is zero. But the zero induced emf doesn't necessarily mean the flux may be zero. However the rate of change of flux is zero (i.e. $ϕ_{B} = c o n s t a n t ⟹ \frac{d ϕ _{B}}{d t} = 0$ )

Figure shows a rectangular conductor PQRS in which the Conductor PQ is free to move in a uniform magnetic field B perpendicular to the plane of the paper. The filed extends from x=0 to x=b and is zero for

x > b

. Assume that only the arm PQ possesses resistance r. when the arm PQ is pulled outward from x=0 to x=2b and is then moved back to x=0 with constant speed v, Determine the expression for the flux and induced emf. Sketch the variations of these quantities with distance

0 \leq x \leq 2 b

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