Electromagnetic Induction Concept Page - 6

Example

Back EMF

When the armature of a D.C. motor rotates under the influence of the driving torque, the armature conductors move through the magnetic field and hence e.m.f. is induced in them as in a generator. The induced e.m.f. acts in opposite direction to the applied voltage  V (Lenzs law) and in known as back or counter e.m.f. Eb$E_b$ 

The back emf Eb=PϕZN/60A$E_b=P\varphi ZN/60A$ is always less than the applied voltage V, although this difference is small when the motor is running under normal conditions

Definition

Relation between self inductance and mutual inductance of 2 coils

Let the number of turns in primary coils are N1$N_1$ and its length is l.The area of cross section of primary coil is A. When the current I flows through it, then flux linked it will be 
ϕ=Magneticfield×effective area$\varphi =Magneticfield\times effectivearea$
ϕ=μoN1Il×N1A$\varphi ={\displaystyle \frac{{\mu }_o{N}_{1}I}{l}}\times N_{1}A$
Self inductance of coil is:
L1=ϕ1I$L_1={\displaystyle \frac{{\varphi }_1}{I}}$
L1=μN21Al$L_1={\displaystyle \frac{\mu N{}_1^{2}A}{l}}$............(1)
Similarly self inductance of secondary coil is 
L2=μN22Al$L_2={\displaystyle \frac{\mu N{}_2^{2}A}{l}}$............(2)
When current I flows through P, then the flux linked coil S is
ϕs=μoN1Il×N2A${\varphi }_s={\displaystyle \frac{{\mu }_o{N}_{1}I}{l}}\times N_{2}A$
Mutual inductance of two coils is 
M=ϕsI$M={\displaystyle \frac{{\varphi }_s}{I}}$
From eq (1) and (2) we get
L1L2−−−−√=μoN1N2Al$\sqrt{L_1{L}_2}={\displaystyle \frac{{\mu }_o{N}_1{N}_{2}A}{l}}$
On comparing this with mutual inductance formula we get 
M=L1L2−−−−√$M=\sqrt{L_1{L}_2}$

Example

Energy stored in an inductor due to a magnetic field

Example:
A long wire carries a current 5 A$5A$. Find the energy stored in the magnetic field inside a volume 1 mm$1mm$3${}^3$ at a distance 10 cm$10cm$ from the wire.Solution:
U$U$ (energy per unit volume)=B22μ0$={\displaystyle {\displaystyle \frac{B^2}{2{\mu }_0}}}$ and 

Energy U=B22μ0×vol.$U={\displaystyle {\displaystyle \frac{B^2}{2{\mu }_0}}\times vol.}$
B=μ0I2πd$B={\displaystyle \frac{{\mu }_{0}I}{2\pi d}}$

U=(μ0I2πd)2×12μ0×vol.$U={\displaystyle {\left({\displaystyle \frac{{\mu }_{0}I}{2\pi d}}\right)}^2\times {\displaystyle \frac{1}{2{\mu }_0}}\times vol.}$
=μ0I28π2d2×vol.$={\displaystyle {\displaystyle \frac{{\mu }_0{I}^2}{8{\pi }^2{d}^2}}\times vol.}$

=4π×10−7×25×10−98×10(10−2)=π8×10−13J$={\displaystyle {\displaystyle \frac{4\pi \times {10}^{-7}\times 25\times {10}^{-9}}{8\times 10({10}^{-2})}}={\displaystyle \frac{\pi }{8}}\times {10}^{-13}J}$

Formula

Self Inductance of plane circular coil

The magnetic field B at centre of coil carrying current i having radiun r and no. of turns as n is given by:
B=μoni2r$B={\displaystyle \frac{{\mu }_{o}ni}{2r}}$
Hence magnetic flux ϕ$\varphi$ is given by 
ϕ=BA=μoniπr22r$\varphi =BA={\displaystyle \frac{{\mu }_{o}ni\pi r^2}{2r}}$
now, setting the value of ϕ$\varphi$, we getL=nϕi$L={\displaystyle \frac{n\varphi }{i}}$
       =μoπn2r2$={\displaystyle \frac{{\mu }_o\pi n^{2}r}{2}}$

Example

Problem based on mutual inductance of different shape

A small square loop of wire of side l$l$ is placed inside a large square loop of side L(L>>l)$L(L>>l)$. If the loops are coplanar and their centres coincide, the mutual induction of the system is directly proportional to
 Suppose outer loops carries a current I$I$. 
field at the center of outer square loop= 2√μIπ(L/2)${\displaystyle \frac{\sqrt{2}\mu I}{\pi (L/2)}}$
as l<<L$l<<L$, flux linkage = MI=4μIπ(L2√)×l2$MI={\displaystyle \frac{4\mu I}{\pi (L\sqrt{2})}}\times l^2$
therefore,  M∝l2L$M\propto {\displaystyle \frac{l^2}{L}}$

Definition

Relationship between voltage and current in an inductor

Any change in the current through an inductor creates a changing flux, inducing a voltage across the inductor. Inductance is determined by how much magnetic field ϕ$\varphi$ through the circuit is created by a given current i
L=ϕi$L={\displaystyle \frac{\varphi }{i}}$
V=dϕdt=d(Li)dt=Ldidt$V={\displaystyle \frac{d\varphi }{dt}}={\displaystyle \frac{d(Li)}{dt}}=L{\displaystyle \frac{di}{dt}}$

Formula

Inductors in series

Ltotal=L1+L2+L3+.....+LN$L_{total}=L_1+L_2+L_3+.....+L_N$
where Ltotal$L_{total}$ is the effective inductance when inductors L1,L2,L3,....,LN$L_1,L_2,L_3,....,L_N$ are connected in series

Formula

Inductors in parallel

• 1Ltotal=1L1+1L2+1L3+....+1LN${\displaystyle \frac{1}{L_{total}}}={\displaystyle \frac{1}{L_1}}+{\displaystyle \frac{1}{L_2}}+{\displaystyle \frac{1}{L_3}}+....+{\displaystyle \frac{1}{L_N}}$

where Ltotal$L_{total}$ is the effective inductance when inductors L1,L2,L3,....,LN$L_1,L_2,L_3,....,L_N$ are connected in parallel

Example

Effective Inductance of Inductors in Series

Example:
A circuit contains two inductors of self-inductance L1$L_1$ and L2$L_2$ in series as shown in the figure. If M$M$ is the mutual inductance, then find the effective inductance of the circuit shown.

Solution:
if a current i$i$ passes through the series combination, 
induced emf in L1,V1=L1didt+Mdidt$L_1,V_1=L_1\frac{di}{dt}+M\frac{di}{dt}$
induced emf in L2,V2=L2didt+Mdidt$L_2,V_2=L_2\frac{di}{dt}+M\frac{di}{dt}$
total induced emf V=(L1+L2+2M)didt

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