Electromagnetic Induction Concept Page - 3

Example

Induced potential difference in a straight conductor due to motion in a uniform magnetic field

A train is moving towards north with a speed of 180$180$ kilometers per hour. If the vertical component of the earths magnetic field is 0.2×10−4$0.2\times {10}^{-4}$ T$T$, the emf induced in the axle of length 1.5$1.5$ m$m$ is:emf = BlV$BlV$
        = 0.2×10−4×1.5×180×10003600$0.2\times {10}^{-4}\times 1.5\times 180\times {\displaystyle \frac{1000}{3600}}$
        =  1.5×10−3V$1.5\times {10}^{-3}V$
        = 1.5$1.5$ mV$mV$

Example

Problem based on potential difference due to rotational motion

A wheel with 10 metallic spokes each 0.5 m long is rotated with a speed of 120 rev/min in a plane normal to the horizontal component of earths magnetic field HE$H_E$ at a place. If HE=0.4G$H_E=0.4G$ at the place, what is the induced emf between the axle and the rim of the wheel? Note that 1G=104T$1G={10}^{4}T$.
Solution
Induced emf=(12)ωBR2=(12)40.4104(0.5)2=6.28×105V$=({\displaystyle \frac{1}{2}})\omega B{R}^2=({\displaystyle \frac{1}{2}})40.4104(0.5)2=6.28\times {10}^{5}V$
The number of spokes is immaterial because the emfs across thespokes are in parallel.

Formula

Force required to move a conductor in a magnetic field

F=I(l×B)=B2l2vr$F=I(l\times B)={\displaystyle \frac{B^2{l}^{2}v}{r}}$
where B is the magnetic field,r is the overall resistance of the loop.
Example: A rectangular loop with a sliding connector CD$CD$ of length l=1.0$l=1.0$ m is situated in uniform and constant magnetic field B=2T$B=2T$ perpendicular to the plane of loop. Resistance of connector CD$CD$ is r=2Ω$r=2\mathrm{\Omega }$. Two resistances of 6Ω$6\mathrm{\Omega }$ and 3Ω$3\mathrm{\Omega }$ are connected as shown in figure. The external force required to keep the connector moving with a constant velocity v=2m/s$v=2m/s$ perpendicular to CD$CD$ and in the plane of the loop is :Since resistances 3 Ω$\mathrm{\Omega }$ and 6 Ω$\mathrm{\Omega }$ are in parallel, tnhe equivalent resistance Req=3×63+6=2Ω$R_{eq}={\displaystyle \frac{3\times 6}{3+6}}=2\mathrm{\Omega }$
Motional emf ε$\epsilon$ through the rod CD$CD$ is ε=Blv$\epsilon =Blv$ where B$B$ is the magnetic field, l$l$ is the length of the rod and v$v$ is the velocity of the rod.
Current I$I$ through the rod is I=BlvReq+r=2×1×22+2=1A$I={\displaystyle \frac{Blv}{R_{eq}+r}}={\displaystyle \frac{2\times 1\times 2}{2+2}}=1A$
Force on the rod F=Bil=2×1×1=2N$F=Bil=2\times 1\times 1=2N$

Formula

Power required to move a conductor in a magnetic field

P=B2l2v2R$P={\displaystyle \frac{B^2{l}^2{v}^2}{R}}$
where B is the magnetic field,
l is the length of the conductor
v is the velocity of the conductor
R is the resistance

Definition

Eddy currents

Currents induced in the conductor due to changing magnetic flux are called eddy currents. They flow in closed loops in plane perpendicular to the magnetic field. The value of eddy currents can be found using faraday's law of electromagnetic induction.

Definition

Applications and disadvantages of eddy currents

Eddy currents are used to advantage in certain applications like:
(i) Magnetic braking in trains: Strong electromagnets are situated above the rails in some electrically powered trains. When the electromagnets are activated, the eddy currents induced in the rails oppose the motion of the train. As there are no mechanical linkages, the braking effect is smooth.
(ii) Electromagnetic damping: Certain galvanometers have a fixed core made of nonmagnetic metallic material. When the coil oscillates, the eddy currents generated in the core oppose the motion and bring the coil to rest quickly.
(iii) Induction furnace: Induction furnace can be used to produce high temperatures and can be utilised to prepare alloys, by melting the constituent metals. A high frequency alternating current is passed through a coil which surrounds the metals to be melted. The eddy currents generated in the metals produce high temperatures sufficient to melt it.
(iv) Electric power meters: The shiny metal disc in the electric power meter(analogue type) rotates due to the eddy currents. Electric currents are induced in the disc by magnetic fields produced by sinusoidally varying currents in a coil.
Disadvantages of eddy current include:
1. Heat is lost in the core of transformers due to eddy currents.
2. The value of eddy current is highly sensitive to the value of permeability. This causes difficulty in the functioning of detectors.

Example

Use of line integral of electric field

Example: A uniform but time-varying magnetic field B(t)$B(t)$ exists in a circular region of radius a$a$ and is directed into the plane of the paper as shown in the figure. Find the magnitude of the induced electric field at point P$P$ at a distance r$r$ form the centre of the circular region.

Solution:
∫E⋅dl=−dΦdt$\int E\cdot dl=-{\displaystyle \frac{d\mathrm{\Phi }}{dt}}$
E⋅2πr=−πa2dBdt$E\cdot 2\pi r=-\pi a^2{\displaystyle \frac{dB}{dt}}$
E=−1r(constant)$E=-{\displaystyle \frac{1}{r}}(constant)$
E∝−1r$E\propto -{\displaystyle \frac{1}{r}}$      (and -ve sign indicates that E$E$ decreases)

Example

Problem in which circuit is rotating in the plane of magnetic field

Example: In a uniform magnetic field of induction B$B$, a wire in the form of semicircle of radius r$r$ rotates about the diameter of the circle with angular velocity ω$\omega$. If the total resistance of the circuit is R$R$, then find the mean power generated per period of rotation.

Solution:
Flux ϕ=BAcosθ=B×(12πr2)cosθ$\varphi =BA\cos \theta =B\times ({\displaystyle \frac{1}{2}}\pi r^2)\cos \theta$
|ε|=|dϕdt|=12πBr2d(cosθ)dt$|\epsilon |=|{\displaystyle \frac{d\varphi }{dt}}|={\displaystyle \frac{1}{2}}\pi B{r}^2{\displaystyle \frac{d(\cos \theta )}{dt}}$
|ε|=12πBr2(sinθ)dθdt$|\epsilon |={\displaystyle \frac{1}{2}}\pi B{r}^2(\sin \theta ){\displaystyle \frac{d\theta }{dt}}$
|ε|=12πBr2(sinωt)ω$|\epsilon |={\displaystyle \frac{1}{2}}\pi B{r}^2(\sin \omega t)\omega$
Power: P=|ε|2R=(12πBr2ωsinωt)2R$P={\displaystyle \frac{|\epsilon {|}^2}{R}}={\displaystyle \frac{({\displaystyle \frac{1}{2}}\pi B{r}^2\omega \sin \omega t{)}^2}{R}}$
Mean Power Pmean=∫T0|ε|2Rdt∫T0dt=π2B2r4ω2∫T0sin2ωtdt4TR$P_{mean}={\displaystyle \frac{{\int }_0^T{\displaystyle \frac{|\epsilon {|}^2}{R}}dt}{{\int }_0^{T}dt}}={\displaystyle \frac{{\pi }^2{B}^2{r}^4{\omega }^2{\int }_0^T{\sin }^2\omega tdt}{4TR}}$
                               =π2B2r4ω24TRT2=π2B2r4ω28R

BookMarks

Page 1  Page 2  Page 3  Page 4  Page 5  Page 6  Page 7  Page 8