Electromagnetic Induction Concept Page - 2

Definition

Faraday's Law of Electromagnetic Induction

According to Faraday's Law of electromagnetic induction:

1. Whenever there is a change in the magnetic flux linked with a coil, an e.m.f. is induced. 
2. The magnitude of e.m.f. induced is directly proportional to the rate of change of magnetic flux linked with the coil.

Note: Magnetic flux through a coil, ϕ=NBA$\varphi =NBA$ where, B$B$ and A$A$ are perpendicular to each other.
|ε|=Δϕ/Δt$|\epsilon |=\mathrm{\Delta }\varphi /\mathrm{\Delta }t$

Law

Lenz's Law

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. Hence, the magnetic field produced by the current in the coil is opposite to the direction of external magnetic field. It is shown by a negative sign in the Faraday's law.
ε=−Δϕ/Δt$\epsilon =-\mathrm{\Delta }\varphi /\mathrm{\Delta }t$

Example

Calculating induced emf due to changing area

A horizontal magnetic field B is produced across a narrow gap between the two square iron pole pieces, A closed square loop of side a. mass m and resistance R is allowed to fall with the top of the loop in the field. The loop attains a terminal velocity equal to:Force on wire il⃗ ×B⃗ $i\overrightarrow{l}\times \overrightarrow{B}$
area ofϕ=xa$\varphi =xa$
dxdt=−V$\frac{dx}{dt}=-V$
ϕ=Bax$\varphi =Bax$
e=−dϕdt=BaV⇒i=BaVR$e=\frac{-d\varphi }{dt}=BaV\Rightarrow i=\frac{BaV}{R}$
the magnetic forces on AD and BC are cancelled by each other. Magnetic field applies upward force on CD wires
F=iBa=B2a2VR$F=iBa=\frac{B^2{a}^{2}V}{R}$
which is equal to mg
B2a2VR=mg$\frac{B^2{a}^{2}V}{R}=mg$
V=RmgB2a2$V=\frac{Rmg}{B^2{a}^2}$

Example

Induced emf due to varying magnetic field

A magnetic field induction is changing in magnitude in a region at a constant rate dBdt${\displaystyle \frac{dB}{dt}}$. A given mass m$m$ of copper drawn into a wire and formed into a loop is placed perpendicular to the field. If the values of specific resistance and density of copper are ρ$\rho$  and σ$\sigma$ respectively, then the induced emf and the resistance in the loop is given by :radius of wire =r$=r$                                           radius of loop=R1$=R_1$
cross-sectional area of wire =πr2$=\pi r^2$               length of wire =2πR1$=2\pi R_1$

Total resistance of wire =R=ρ×2πR1πr2$=R=\rho \times {\displaystyle \frac{2\pi R_1}{\pi r^2}}$
mass=m,density=σ$=m,density=\sigma$
σ×πr2×2πR1=m$\sigma \times \pi r^2\times 2\pi R_1=m$
∴R=ρ×(2πR1)2σm$\therefore R=\rho \times (2\pi R_1{)}^2{\displaystyle \frac{\sigma }{m}}$
emf=dBdt×πR21$emf={\displaystyle \frac{dB}{dt}}\times \pi R_1^2$

Example

Induced emf in a coil caused by varying flux due to magnetic field by a current carrying coil

In the shown figure, there are two long fixed parallel conducting rails (having negligible resistance) and are separated by distance L$L$. A uniform rod(cd) of resistance R$R$ and mass M$M$ is placed at rest on frictionless rails. Now at time t=0,$t=0,$ a capacitor having charge Q0$Q_0$ and capacitance C$C$ is connected across rails at ends a and b such that current in rod(cd) is from c towards d and the rod is released. A uniform and constant magnetic field having magnitude B$B$ exists normal to plane of paper as shown. (Neglect acceleration due to gravity)At any instant t,$t,$ the charge on capacitor q$q$, velocity of rod v$v$ and the current I$I$ through rod are as shown

∴mdvdt=BIL=B(−dqdt)L....(1)$\therefore m{\displaystyle \frac{dv}{dt}}=BIL=B(-{\displaystyle \frac{dq}{dt}})L....(1)$
⇒∫v0mdv=−∫qQ0BLdq$\Rightarrow {\int }_0^{v}mdv=-{\int }_{Q_0}^{q}BLdq$
solving we get q, q=Q0−MvBL................(2)$q,q=Q_0-{\displaystyle \frac{Mv}{BL}}................(2)$
AlsoqC=BLV+IR=BLv−RdqBL..........(3)${\displaystyle \frac{q}{C}}=BLV+IR=BLv-R{\displaystyle \frac{dq}{BL}}..........(3)$
From equations (1)$(1)$ and (3)$(3)$
qC=Blv+mRBLdvdt.................(4)${\displaystyle \frac{q}{C}}=Blv+{\displaystyle \frac{mR}{BL}}{\displaystyle \frac{dv}{dt}}.................(4)$
from (4) when dvdt=0⇒qC=Blv....(5)${\displaystyle \frac{dv}{dt}}=0\Rightarrow {\displaystyle \frac{q}{C}}=Blv....(5)$
From (2) and (5). At instant acceleration is zero
v=Q0LBM+B2L2C$v={\displaystyle \frac{Q_{0}LB}{M+B^2{L}^{2}C}}$ and q=B2L2CQ0M+B2L2C$q={\displaystyle \frac{B^2{L}^{2}C{Q}_0}{M+B^2{L}^{2}C}}$

Definition

Direction of electric current in loops using Lenz's law

When the North-pole of a bar magnet is moved towards a closed loop like a coil connected to a galvanometer, the magnetic flux through the coil increases. Hence according to Lenz's law, current is induced in the coil in such a direction that it opposes the increase in flux.This is possible only if the current in the coil is in a counter-clockwise direction with respect to an observer situated on the side of the magnet.Note that magnetic moment associated with this current has North polarity towards the North-pole of the approaching magnet. Similarly, if the North pole of the magnet is being withdrawn from the coil, the magnetic flux through the coil will decrease. To counter this decrease in magnetic flux,the induced current in the coil flows in clockwise direction and its South pole faces the receding North-pole of the bar magnet. This would result in an attractive force which opposes the motion of the magnet and the corresponding decrease in flux.

Result

Factors affecting induced current

Induced current in the coil can be increased by increasing the number of turns in the coil, rapidly moving the coil and increasing the applied magnetic field.

Definition

Motional emf

When a metal rod of length l$l$ is placed normal to a uniform magnetic field B$B$ and moved with a velocity v$v$ perpendicular to the field, the induced emf (called motional emf) across its ends is
ϵ=Blv

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