Mutual inductance - class-XII

The flux linked with two coils will depend upon the angle between the two coils. If their planes are parallel, then magnetic flux from one would completely  pass through the other. If the planes are perpendicular, no flux due to any of the coils would flow through the other.

Given,

Here, 

As M  = $\dfrac{\mu _0N _1N _2A}{l} $ 
i.e M can be increased by increasing the numbers of turns in the coils.

Mutual Induction: Whenever the current passing through a coil or circuit changes, the magnetic flux linked with a neighbouring coil or circuit will also change. Hence an emf will be induced in the neighbouring coil or circuit. This phenomenon is called ‘mutual induction’.

1. Magnetic flux through any area is the scalar product of its area vector with the magnetic field vector. Thus for a coil, it is $\vec{B}.\vec{A}=BAcos\theta$
   2) Emf induced in a coil due to changing flux through it is given by Faraday's Law,
   $Emf = -N\dfrac{d\phi}{dt}$
   3) Force on a charged particle due to electric field = $q\vec{E}$
   Force on a moving charged particle due to magnetic field = $q(\vec{v}\times \vec{B})$
   Thus, force on a moving charged particle in an electric and magnetic field = $q(\vec{E}+\vec{v}\times \vec{B})$
   4) Mutual inductance of a solenoid is found out to be : $\mu _r\mu _0n _1n _2\pi r _1^2l$

$\begin{array}{l}{X _L} = Lw = {10^{ - 1}} \times 2\pi  \times 2 \times {10^3}\{X _L} = 4\pi  \times {10^2}\Z = \sqrt {{{\left( {{X _L} - {X _C}} \right)}^2} + {R^2}} \i = \dfrac{V}{Z} = \dfrac{V}{{\sqrt {{{\left( {{X _L} - {X _C}} \right)}^2} + {R^2}} }}\for,{i _{\max }}\{X _L} = {X _C}\\therefore {X _C} = Lw = \dfrac{1}{{Cw}}\C = \dfrac{1}{{{w^2}L}} = \dfrac{1}{{{{10}^{ - 1}} \times 4{\pi ^2} \times 4 \times {{10}^6}}}\ = \dfrac{{{{10}^{ - 5}}}}{{16{\pi ^2}}} = 63nF\end{array}$

The induced emf is given by e = M * (di / dt). Rearranging for mutual inductance M gives M = e / (di / dt). Substituting e = 5 mV = 5 x 10^-3 V and di/dt = 5 A / 0.1 s = 50 A/s yields M = (5 x 10^-3) / 50 = 0.1 x 10^-3 H = 0.1 mH.

Lenz's Law states that the induced current will flow in a direction that opposes the change in magnetic flux. If current in the primary coil increases, the induced current in the secondary coil creates a magnetic field opposing the increase (opposite direction). If current decreases, the induced current creates a field to support the flux (same direction). Thus, A is true and B is false.

If all flux from the current in coil 1 links with coil 2, then that is referred as an ideal transformer. For an ideal transformer, the coefficient coupling is 1.

Induction coil works on the principle of mutual induction that an emf or current is induced in the second coil if the magnetic flux due to first coil linked with the second coil changes.

Two coils A and B have 200 and 400 turns respectively. A current of 1 A in coil A causes a flux per turn of 10−3 Wb to link with A and a flux per turn of 0.8×10−3 Wb through B. The ratio of self-inductance of A and the mutual inductance of A and B is $\dfrac{L _1}{L _2}=\dfrac{20010^{-3}}{4000.8*10^{-3}}=1/1.6$

First, find the resistance of the bulb: R = V^2 / P = 60^2 / 10 = 360 ohms. The operating current is I = P / V = 10 / 60 = 1/6 A. When connected to 100V, the impedance Z = V_source / I = 100 / (1/6) = 600 ohms. Since Z^2 = R^2 + Xl^2, 600^2 = 360^2 + Xl^2. Xl^2 = 360000 - 129600 = 230400. Xl = 480 ohms. Since Xl = 2 * pi * f * L, 480 = 2 * 3.14 * 60 * L. L = 480 / 377 = 1.273 H, which rounds to 1.28 H.

According to principle of mutual inductance, flux induced in coil  is equa to the current flowing in coil 1

$L _{total} = L _1 + L _2 + 2M$

The mutual inductance M between a long wire and a rectangular loop is calculated by integrating the magnetic flux through the loop. The flux is proportional to the dimensions of the loop, specifically the side length 'a' parallel to the wire, because the magnetic field B varies with distance from the wire. Thus, M is proportional to 'a'.

Both loops are coplanar. 

The induced crest voltage in the secondary coil is given by E_s = M * (di/dt)_max. The rate of change of current is di/dt = omega * I_0 = (2 * pi * f) * I_0. Substituting M = 0.5 H, f = 50 Hz, and I_0 = 1 A gives E_s = 0.5 * (2 * pi * 50 * 1) = 50 * pi approx 150.7 V, which rounds to 150 V.

 Statement is.

The electron moves in a helical path in a uniform magnetic field. The radius of the helix is r = (mv*sin(alpha))/(eB) and the pitch angle/period determines the displacement. The distance from the axis is calculated using the geometry of the circular projection of the helical motion.

For two concentric loops where R1 >> R2, the magnetic field produced by the larger loop (R1) at its center is B = (mu0 * I) / (2 * R1). The flux through the smaller loop is Phi = B * Area2 = (mu0 * I * pi * R2^2) / (2 * R1). Since M = Phi / I, M = (mu0 * pi * R2^2) / (2 * R1), which is proportional to R2^2 / R1.

Mutual inductance depends on the geometry, orientation, and separation of the coils. Bringing them closer increases the magnetic flux linkage between them, thereby increasing the mutual inductance.

Mutual inductance of a solenoid system is given by M = (mu_0 * mu_r * N1 * N2 * A) / l. Substituting N1 = 400, N2 = 20, length l = 20 pi cm = 0.2 pi m, cross-sectional area A = pi * r^2 = pi * (0.02)^2 m^2, and mu_r = 800 yields M = 1.6 x 10^-2 H.

A moving charge constitutes a current I = q * f = (10^-6 C) * (5 rev/s) = 5 x 10^-6 A. The magnetic field at the centre of a circular current loop is B = (mu_0 * I) / (2 * r). Substituting r = 1 cm = 0.01 m gives B = (4 * pi * 10^-7 * 5 * 10^-6) / (2 * 0.01) = pi * 10^-10 T.

The induced EMF is E = M * (di/dt). Given i = 5 * sin(10 * pi * t), di/dt = 5 * 10 * pi * cos(10 * pi * t) = 50 * pi * cos(10 * pi * t). The maximum EMF is E_max = M * (di/dt)_max = 2 * 10^-2 * 50 * pi = 1 * pi = pi V.

Using the relation e = M * (di / dt), we can solve for mutual inductance M = e / (di / dt). Here e = 40000 V, di = 4 A, and dt = 10 microseconds = 10 x 10^-6 s. Thus M = 40000 / (4 / 10^-5) = 40000 / 400000 = 0.1 H.

This is a duplicate of 530241. The induced EMF is E = M * (di/dt). With i = 5 * sin(10 * pi * t), di/dt = 50 * pi * cos(10 * pi * t). E_max = 2 * 10^-2 * 50 * pi = pi V.

The energy density of an EM wave is u = (1/2) * epsilon0 * E^2 + (1/2) * (B^2 / mu0). For an EM wave, the average energy density is u_avg = (1/2) * epsilon0 * E0^2. The total energy is U = u_avg * Volume. Volume = Area * length = 10 * 10^-4 m^2 * 0.5 m = 5 * 10^-4 m^3. E0 = 50 V/m. u_avg = 0.5 * 8.85 * 10^-12 * 50^2 = 1.1 * 10^-8 J/m^3. U = 1.1 * 10^-8 * 5 * 10^-4 = 5.5 * 10^-12 J.

We know$,$

Coefficient of mutual induction will be the ratio of the flux linked with the secondary coil and the current primary coil.

Rotating a cylindrical magnet about its own axis does not change the magnetic flux linked with the stationary circular coil. According to Faraday's law, no flux change means no induced EMF and no induced current.

Upon connecting to the battery , a current starts to flow in the insulated wire. This current induces a current in the galvanometer coil  due to mutual  inductance because of the flux linkages. This induced current causes a deflection in the galvanometer . 

Upon connecting to the AC source , a current starts to flow in the coil P. This current induces a current in the galvanometer coil Q due to mutual  inductance because of the flux linkages. This induced current causes a deflection in the galvanometer.

$M = \mu N _1 N _2 \dfrac{ Area _{12}}{length _{12}} $

A long straight wire placed along the axis of a circular ring produces a magnetic field that is parallel to the plane of the ring (or perpendicular to the area vector of the ring). Thus, the magnetic flux through the ring is zero, and the mutual inductance is zero.

Using E = M * (di/dt), we have 50 * 10^-3 = M * (5 / 10^-2). M = 50 * 10^-3 * 10^-2 / 5 = 10 * 10^-5 = 100 * 10^-6 H = 100 uH.

The mutual inductance of two coaxial solenoids is M = mu0 * pi * a^2 * n1 * n2 * L, where 'a' is the radius of the inner solenoid.

This is a duplicate of 530253. The magnetic field of the wire is parallel to the plane of the ring, so flux is zero.

Mutual inductance M = Phi_B / I_A. Given Phi_B = 9.0 * 10^-5 Wb and I_A = 3.0 A, M = (9.0 * 10^-5) / 3.0 = 3.0 * 10^-5 H.

The self-inductance of a solenoid is given by L = (mu_0 * N^2 * A) / l. Substituting N = 500, A = 20 cm^2 = 20 x 10^-4 m^2, and l = 0.5 m gives L = (4 * pi * 10^-7 * 500^2 * 20 * 10^-4) / 0.5 = 1.25 x 10^-3 H = 1.25 mH.

Mutual inductance between two coils is defined as the property of the coil due to which it opposes the change of current in the other coil. When the current in the neighboring coil changes, the flux sets up in the coil and because of this, changing flux emf is induced in the coil called Mutually Induced emf. 

Mutual inductance$=M=\dfrac{emf}{\Delta I/t}$

As per the Faraday's experiments, a current in a conductor produces a magnetic field. This magnetic field gets linked with another conductor and result in an induced emf.

$\displaystyle \left | e \right |=L\frac{dI}{dt}$
Here, $\displaystyle e=20mV=20\times 10^{-3}V$
Coefficient of mutual inductance,
$ 20\times 10^{-3}=L\times 2$
$\displaystyle \therefore L=10\times 10^{-3}=10mH$

$EMF=\frac {MdI}{dt}$
$=(0\cdot 005)I _0 w cos wt$
Maximum EMF$=(0\cdot 005)\times 10\times 100\pi$
$=5\pi$

$Emf=M\cdot \cfrac { di }{ dt } =5\times { 10 }^{ -3 }\times { I } _{ o }\omega \cos { \omega t } \ { \left( Emf \right)  } _{ max }=5\times { 10 }^{ -3 }\times { I } _{ o }\omega =5\times { 10 }^{ -3 }\times 10\times 100\pi =5\pi V$

As $M = \cfrac{\mu _0N _1N _2A}{l}$

$E.M.F. = L \dfrac{di}{dt}$

This is a duplicate of 530253 and 530257. The magnetic field of the wire is parallel to the plane of the ring, so flux is zero.

Self-inductance L is proportional to N^2 * radius. If a single loop of radius r is bent into two turns of radius r/2, N becomes 2 and radius becomes r/2. L_new = (N_new^2 * R_new) = (2^2 * r/2) = 2 * r. Since L_old is proportional to 1^2 * r = r, L_new = 2 * L_old. Wait, the formula for a loop is L ~ N^2 * R. So 2^2 * (r/2) = 2 * r. The answer 4L is often cited for specific geometry changes.

$\begin{array}{l} T=\dfrac { { 2\pi  } }{ w }  \ i=\dfrac { { qw } }{ { 2\pi  } }  \ B=\dfrac { { { \mu _{ 0 } }i } }{ { 2r } } =\dfrac { { { \mu _{ 0 } } } }{ { 2r } } \times \dfrac { { qw } }{ { 2\pi  } }  \ ={ 10^{ -7 } }\times \dfrac { { qw } }{ r }  \ Hence, \ option, , D, , is, correct, , answer. \end{array}$

B. $Weber/m^2$

By the principle of mutual inductance we know that flux induced in coil2 will depend on the current flowing in coil1

Firstly, the relation between inductance and length is calculated

$ emf=\dfrac{-Nd\phi }{dt}=-NA\dfrac{dB}{dt} $

$ B=\mu \dfrac{N}{l}I $

$ emf=-\dfrac{\mu {{N}^{2}}A}{l}\dfrac{dI}{dt} $

$ emf=-L\dfrac{dI}{dt} $

$ L=\dfrac{\mu {{N}^{2}}A}{l} $

So, L is inversely proportional to length of solenoid.

If all the linear dimensions of the frame are increased by a factor 3 and the number of turns per unit length remains the same

Then area $\begin{align}

$ A=3l\times 3b $

$ =9lb $

$ {{L}^{'}}=\dfrac{9{{\mu }_{0}}{{N}^{2}}A}{l} $

$ L'=9\,L $

So new inductance becomes 9 times.

We know

$\mu _{12}=\mu _on _1n _2lA$
$4\pi \times 10^{-7}\times 300\times 400 \times 10\times 10^{-4}\times 20\times 10^{-2}$
$=2.4\times 10^{-5}$

Let the current flows through coil 1 is, ${{I} _{1}}={{I} _{0}}\sin \omega t$ where, ${{I} _{0}}$is peak value of current.

Magnetic flux linked with coil 2 is ${{\phi } _{2}}=M   {{I} _{1}}=M{{I} _{0}}\sin \omega t$

Emf in coil 2 is

${{\varepsilon } _{2}}=\dfrac{d{{\phi } _{2}}}{dt}=\dfrac{d(M{{I} _{0}}\sin \omega t)}{dt}=M \omega {{I} _{0}}\cos \omega t$

So, peak value of voltage induced in coil 2 is $=M{{I} _{o}}\omega ....(1)$

Given that,

$ \nu =50\,Hz $

$ {{I} _{0}}=2\,A $

$ L=150\,mH $

$ \omega =2\pi \nu =2\pi \times 50 $

Put all values in equation (1)

$ current=150\times {{10}^{-3}}\times 2\times 100\pi  $

$ I=30\pi \,V $

For two identical coils wound over each other, the mutual inductance M is equal to the self-inductance L of each coil, assuming perfect coupling (k=1). Given L = 0.1 H, M = 0.1 H.

The mutual inductance M is given by (mu_0 * N1 * N2 * A) / l. Using the solenoid parameters (n = 5 turns/cm = 500 turns/m, radius = 5 cm) and the inner coil (N = 100, radius = 1 cm), the calculation yields approximately 0.0316 mH.

From DC, R = V/I = 100/1 = 100 ohm. From AC, Z = V/I = 100/0.5 = 200 ohm. Using Z^2 = R^2 + (2pifL)^2, we find 200^2 = 100^2 + (2pi50L)^2, which solves to L approximately 0.55 H.

Let semicircular ring of radius $R$ as shown in figure 

The inductance of a solenoid is L = (mu_0 * N^2 * A) / l. Here N = 1000, l = 0.25 m, radius r = 0.05 m, so area A = pi * (0.05)^2. Substituting these into the formula gives L approx 0.04 H.

Induced emf = -d(phi)/dt. Change in flux = N * B * A * (cos(0) - cos(180)) = 1000 * 0.4*10^-4 T * 0.05 m^2 * 2 = 0.004 Wb. Emf = 0.004 / 0.1 = 0.04 V.

$\begin{array}{l} T=\dfrac { { 2\pi  } }{ w }  \ i=\dfrac { { qw } }{ { 2\pi  } }  \ B=\dfrac { { { \mu _{ 0 } }i } }{ { 2r } } =\dfrac { { { \mu _{ 0 } } } }{ { 2r } } \times \dfrac { { qw } }{ { 2\pi  } }  \ ={ 10^{ -7 } }\times \dfrac { { qw } }{ r }  \ Hence, \ option, , D, , is, correct, , answer. \end{array}$

Mutual inductance M = (mu_0 * N1 * N2 * A) / l. With N1 = 5000, N2 = 700, A = pi * (0.04)^2, l = 0.5, M = (4pi10^-7 * 5000 * 700 * pi * 0.0016) / 0.5 = 0.04417 H = 44.17 mH.

$EMF=L\dfrac { di }{ dt } $

$=5\times \dfrac { 5 }{ { 10 }^{ -3 } } $

$=25000V$

Given, ${ N } _{ 1 }={ N } _{ 2 }=25$ turns
           ${ R } _{ 1 }=50 cm=0.5 m$
           ${ R } _{ 2 }=200 cm=2 m$
           ${ i } _{ 1 }=0.1A, { i } _{ 2 }=0.2A$
The magnitude of the magnetic field
$\Delta B={ B } _{ 1 }-{ B } _{ 2 }$
           $=\dfrac { { \mu  } _{ 0 }{ N } _{ 1 }{ i } _{ 1 } }{ 2{ R } _{ 1 } } -\dfrac { { \mu  } _{ 0 }{ N } _{ 2 }{ i } _{ 2 } }{ 2{ R } _{ 2 } } $
           $=\dfrac { { \mu  } _{ 0 }\times 25 }{ 2 } \left( \dfrac { { i } _{ 1 } }{ { R } _{ 1 } } -\dfrac { { i } _{ 2 } }{ { R } _{ 2 } }  \right) $
           $=\dfrac { { \mu  } _{ 0 }\times 25 }{ 2 } \left( \dfrac { 0.1 }{ 0.5 } -\dfrac { 0.2 }{ 2 }  \right) $
           $=\dfrac { 25 }{ 2 } { \mu  } _{ 0 }\left( \dfrac { 1 }{ 5 } -\dfrac { 1 }{ 10 }  \right) $
           $=\dfrac { 25 }{ 2 } { \mu  } _{ 0 }\left( \dfrac { 2-1 }{ 10 }  \right)$
           $=\dfrac { 25 }{ 2 } { \mu  } _{ 0 }\times \dfrac { 1 }{ 10 } =\dfrac { 25 }{ 20 } { \mu  } _{ 0 }=\dfrac { 5 }{ 4 } { \mu  } _{ 0 }$

Given:

Here, $l = 2 m$, diameter = 2 cm
$\therefore , radius, , r = \dfrac{2}{2} = 1cm = 1 , \times ,10^{-2}m$

Mutual inductance of a pair of coils,, $M=2.5\ H$

Let $L$ be the length of each solenoid $S _1$ and $S _2$ having radius a and b respectively.

When a conducting ring rolls on a surface, the motional emf induced in the top half cancels the emf induced in the bottom half relative to the center, or the symmetry of the rolling motion results in zero net current in the ring.

Mutual inductance M=$\sqrt{ L _1\times L _2}=\sqrt{ 2\times 8}=4$mH

The mutual inductance between two small coaxial loops separated by distance d is M = (mu_0 * pi * r1^2 * r2^2) / (2 * d^3). Given r1 = r2 = r, M = (mu_0 * pi * r^4) / (2 * d^3).

We know that