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Instantaneous Value Of Induced Emf

Learn how to determine the magnitude of induced EMF due to the flux in the coil and find the maximum induced EMF in the coil and the circuit. Find the correct graph of induced EMF, if loop starts entering at t = 0 sec.

Graph of EMF from the Graph of Flux

  • When flux is given as a function of time \(i.e.\,\,\,\,\phi(t)=f(t)\), as shown in graph.

\(\phi(t)=f(t)\) is a equation of straight line.

  • Induced EMF 

The slope of the curve represents the induced emf as \(\mathcal{E}=-\dfrac{d\phi(t)}{dt}\)

The graph of derivative of \(\phi(t)\) gives the graph of emf with time, which is shown as follows

  • Example :

let  \(\phi(t)=2t +3\)

slope of this equation will be 

\(m=\dfrac{d\phi(t)}{dt}=\dfrac{d}{dt}(2t+3)\)

\(m=2\)

thus,  \(\mathcal{E}=-2V\)

The graph of emf with time will be - 

Illustration Questions

The graph of flux against time is shown in figure. Which of the following graph represents emf?

A

B

C

D

×

Slope of graph from 0 to 1 sec :

\(\dfrac{d\phi}{dt}=m=\dfrac{2-0}{1-0}=2\)

Thus,  \(emf=-\dfrac{d\phi}{dt}=-2V\)

image image

Slope of graph from 1 to 2 sec :

\(\dfrac{d\phi}{dt}=m=0\)

Thus,  \(emf=-\dfrac{d\phi}{dt}=0V\)

image image

Slope of graph from 2 to 3 sec :

\(\dfrac{d\phi}{dt}=m=\dfrac{0-2}{3-2}=-2\)

Thus,  \(emf=-\dfrac{d\phi}{dt}=-(-2)=+2V\)

image image

Hence, option (D) is correct.

image

The graph of flux against time is shown in figure. Which of the following graph represents emf?

image
A image
B image
C image
D image

Option D is Correct

Induced EMF in the Circuit

  • Consider a circuit in which a conducting rod of length  \(\ell\) is placed over two rail which are clamped as shown in figure.
  • The region contains uniform magnetic field \(\vec B\) and the rod is moving with velocity \(v\).
  • Since, the rod is moving with velocity \(v\), therefore all the charges moving in rod also move with velocity \(v\).

  • Due to the movement of rod, the positive charge will experience force in upward direction while negative charge will experience force in downward direction.
  • Due to the force experienced by charges, the charges get accumulated, as shown in figure.
  • The accumulation of charges creates an electric field whose direction is along (–y)–axis. Due to this field, the electric force also generates.
  • The charge carriers will continue to move until the electric force \(F_E=qE\) exactly balance the magnetic force, \(F_B=qvB\)
  • The balance condition is

\(F_B=F_E\)

\(qvB=qE\)

\(E=vB\)

 

 

  • The potential difference across the ends of rod :

\(\Delta V=E\ell\)

\(\because\,E=vB\)

\(\therefore\,\Delta V=Bv\ell\)

  • Magnetic flux :

\(\phi=B\ell x\)

  • Emf in the circuit :

\(\dfrac{d\phi}{dt}=B\ell\dfrac{dx}{dt}\)

\(\dfrac{dx}{dt}=v\), where \(v\) is velocity of rod)

\(\dfrac{d\phi}{dt}=B\ell v \)

\(\because\,\,emf=\dfrac{-d\phi}{dt}\)

\(\therefore\)  The induced emf is,

\(\mathcal{E}=-B\ell v\)

Illustration Questions

In the shown figure, the rod of length \(\ell=1m\) is moving with velocity \(v=2m/s\) in the magnetic field, \(B=1\,T\). Find the magnitude of induced emf in the circuit. 

A 2 V

B 3 V

C 4 V

D 5 V

×

Length of rod, \(\ell=1\,m\)

Velocity of rod, \(v=2\,m/s\)

Magnetic field, \(B=1\,T\)

Magnitude of flux is given by

\(|\mathcal{E}|=B\ell v\)

\(|\mathcal{E}|=(1)(1)(2)\)

\(|\mathcal{E}|=2\,V\)

In the shown figure, the rod of length \(\ell=1m\) is moving with velocity \(v=2m/s\) in the magnetic field, \(B=1\,T\). Find the magnitude of induced emf in the circuit. 

image
A

2 V

.

B

3 V

C

4 V

D

5 V

Option A is Correct

Emf when Alignment of Surface Changes with Time

  • Consider any surface whose area is A and is placed in uniform magnetic field \(\vec B\).
  • The alignment of surface changes with time \(i.e.\), the angle between \(\vec A\) and magnetic field vector \(\vec B\) is varied as a function of time \(\theta=f(t)\). The area of surface and magnetic field is constant.
  • Magnetic flux :

\(\phi _m=\vec B.\,\vec A\)

\(\phi _m=BA\,cos\theta\)

\(\because\,\theta=[f(t)]\)

\(\therefore\,\phi_m=BA \,cos[f(t)]\)

  • Magnitude of Emf in the surface :

\(|\mathcal{E}|=\dfrac{d\phi_m}{dt}\)

\(|\mathcal{E}|=\dfrac{d}{dt}\{BA \,cosf(t)\}\)

 

\(|\mathcal{E}|=BA\,\dfrac{d}{dt}\{cosf(t)\}\)

Illustration Questions

A ring of radius \(r=1\,m\) is placed in a uniform magnetic field \(\vec B=1\,T\), such that the angle between area vector and magnetic field is varied as a function of time, \(\theta=2\,t+3\). Find the magnitude of induced emf at \(t=1\,sec.\)  \([sin5^\circ=0.087]\) Here, \(\theta\)is in degrees.

A 0.948 V

B 0.546 V

C 0.847 V

D 0.666 V

×

Radius of ring,   \(r=1\,m\)

Magnetic field,  \(\vec B=1\,T\)

Angle,  \(\theta=2\,t+3\)

Time,  \(t=1\,sec\)

The magnetic flux is given by 

\(\phi_ m=|\vec B. \,\vec A|\)

\(\phi _m=|B A \,cos\theta|\)

\(\phi_ m=|B A \,cos(2\,t+3)|\)

The induced emf is given by

\(\mathcal{E}=-\dfrac{d\phi}{dt}=|-2BA\,sin (2t+3)\,|\)

\(\mathcal{E}=|-2\times1\times (\pi (1)^2)sin (2\times1+3)|\)

\(\mathcal{E}=2\pi \,sin\,5^\circ\)

\(\mathcal{E}=2\pi (0.087)\)

\(\mathcal{E}=0.546\,V\)

A ring of radius \(r=1\,m\) is placed in a uniform magnetic field \(\vec B=1\,T\), such that the angle between area vector and magnetic field is varied as a function of time, \(\theta=2\,t+3\). Find the magnitude of induced emf at \(t=1\,sec.\)  \([sin5^\circ=0.087]\) Here, \(\theta\)is in degrees.

A

0.948 V

.

B

0.546 V

C

0.847 V

D

0.666 V

Option B is Correct

Graph of EMF Linked with a Loop Entering and Leaving the Magnetic Field

  • Consider a magnetic field region of width 3b.
  • In this region, a rectangular loop of side  \(\ell\)  and width \(b\) enters with uniform velocity \(v\), as shown in figure.

Case I :

When loop is entering in magnetic field from 0 to b width of loop :

  • The loop starts entering at t = 0 and it fully enters at t = t1.
  • While entering, the area of loop which is in contact with \(\vec B\) increases. Hence, the flux increases upto the width of loop.
  • If \(\phi\) is positive, then \(\dfrac{d\phi}{dt}\) will also positive.
  • \(Emf,\,\mathcal{E}=-\dfrac{d\phi}{dt}\)

\(\mathcal{E}=-B\ell v\)

 

 

  • Graph of flux and emf with time, when loop is entering in region of \(\vec B\).

Case II : 

When loop fully enters in region of magnetic field :

  • The loop fully enters at t = t1 and at t = t2 , it just starts leaving the region of \(\vec B\).
  • Here, whole area of loop remains in contact with magnetic field before leaving.
  • Thus, there is no change in flux.

\(\phi=0\)

\(\Rightarrow\,\,\dfrac{d\phi}{dt}=0\)

\(\Rightarrow\,\,\mathcal{E}=0\)

 

 

  • Graph of flux and emf with time, when the loop remains in constant with \(\vec B\).

Case III :

When the loop fully left the region of \(\vec B\) :

  • The loop starts leaving the region at t = t2 and fully left the region of \(\vec B\) at t = t3.
  • While leaving the region of \(\vec B\), the flux through the loop decreases, thus, \(\dfrac{d\phi}{dt}\) is negative.

Hence, \(\mathcal{E}=\dfrac{-d\phi}{dt}=B\ell v\)

Emf is positive.

 

  • Graph of flux and emf with time, when the loop leaves the region of \(\vec B\).

  • Combining the graph of flux and emf of all the three cases :

\(\phi(t)\) with time :

\(\mathcal{E}\) with time :

Illustration Questions

A rectangular loop of dimensions \(\ell\times b=2.0\times 3.0 \,m^2\) enters in a uniform magnetic field, \(B=1\,T\) whose region is of width \(w=9.0\,m\), with uniform velocity, \(v=2\,m/s\), as shown in figure. Choose the correct graph of induced emf, if loop starts entering at t = 0 sec.

A

B

C

D

×

Dimensions of loop, \(\ell\times b=(2\times3)\,m^2\)

Magnetic field, \(B=1\,T\)

width of magnetic field region, \(\text w=9\,m\)

Velocity of loop, \(v=2\,m/s\)

Time taken by loop to enter fully in the region.

t= \(\dfrac{\text distance}{\text speed}\)

t1\(\dfrac{3}{2}\)

t1 = 1.5 sec

Time taken by loop to start leaving the region,

t\(\dfrac{\text distance}{\text speed}\)

t2 =  \(\dfrac{9}{2}\)

t2 = 4.5 sec

Time taken by loop to fully left the region,

t= \(\dfrac{\text distance}{\text speed}\)

t3 = \(\dfrac{12}{2}\)

t3 = 6 sec

As rectangular loop enters in \(\vec B\), flux increases till t1.

The flux remain constant from t1 to t2. The flux starts decreasing from t2 till t3 as the loop leaves \(\vec B\)

The graph of flux will be as follows :

image

Induced emf is given by,

\(\mathcal{E}=B\ell v\)

\(\mathcal{E}\) \(= 1×2×2\)

\(\mathcal{E}\) \(= 4\,V\)

For the time interval, 0 - t1 sec

 \(\dfrac{d\phi}{dt}\) is positive.

As , \(\mathcal{E}=-\dfrac{d\phi}{dt}\)

\(\Rightarrow\,\,\,\mathcal{E}\) is negative.

For the time interval, t- t2 sec,

 \(\dfrac{d\phi}{dt}\) is zero.

\(\Rightarrow\,\,\,\mathcal{E}\) is zero.

For the time interval, t- tsec,

 \(\dfrac{d\phi}{dt}\) is negative.

\(\Rightarrow\,\,\,\mathcal{E}\) is positive.

Thus, the graph of emf with time will be as follows :

image

A rectangular loop of dimensions \(\ell\times b=2.0\times 3.0 \,m^2\) enters in a uniform magnetic field, \(B=1\,T\) whose region is of width \(w=9.0\,m\), with uniform velocity, \(v=2\,m/s\), as shown in figure. Choose the correct graph of induced emf, if loop starts entering at t = 0 sec.

image
A image
B image
C image
D image

Option A is Correct

Induced EMF in a Rotating Coil

  • Consider a coil of area A rotating in a magnetic field.
  • Due to rotation of coil, the magnetic flux through the area enclosed by coil changes with time.
  • The change in magnetic flux induces an emf and current in the coil.
  • Suppose, there are N turns of the coil having same area, A.
  • The coil rotates with constant angular velocity \(\omega\).
  • Flux through the coil at any time t :

If  \(\theta\) is the angle between \(\vec A\) and \(\vec B\) then

\(\phi _m=BA\,cos\theta\)

\(\because\,\theta=\omega t\)

\(\therefore\,\phi _m=BA\,cos\,\omega t\)  

  • Induced emf in the coil :

since,  \(\mathcal{E}=-N\dfrac{d\phi_m}{dt}\)

Thus,  \(\mathcal{E}=-N\dfrac{d}{dt}(BA\,cos \,\omega t)\)

\(\mathcal{E}=-NBA\,(-sin \,\omega t)\omega\)

\(\mathcal{E}=NBA\,\omega \,sin \,\omega t\)

When \(\omega t\) becomes \(90^\circ\), then, \(\mathcal{E}\) will be maximum.

\(\mathcal{E}_{max}=NBA\,\omega \)

Generators and Motors :

  • To produce electricity, same concept is used in the functioning of generator and motor.
  • The loop present in them is mechanically rotated by external rotator in uniform magnetic field. Due to the rotation, emf is induced into the loop to produce electricity.

Illustration Questions

A coil of a wire having \(N=5\) turns of area \(A=2\,cm^2\), rotates in magnetic field of \(\vec B=1\,T\) at constant frequency of \(f=60\,Hz.\) Find the maximum induced emf in the coil.

A 2 V

B 0.2 V

C 0.37 V

D 0.30 V

×

Number of turns in coil, \(N=5\)

Area of each turn, \(A=2\,cm^2\)

Magnetic field,  \(\vec B=1\,T\)

Frequency, \(f=60\,Hz\)

Maximum induced emf is given by

\(\mathcal{E}_{max}=NBA\,\omega \)

\(\mathcal{E}_{max}=5\times2\times10^{-4}\times1\times2\pi\times60\)

\(\mathcal{E}_{max}=0.37\,V\)

A coil of a wire having \(N=5\) turns of area \(A=2\,cm^2\), rotates in magnetic field of \(\vec B=1\,T\) at constant frequency of \(f=60\,Hz.\) Find the maximum induced emf in the coil.

A

2 V

.

B

0.2 V

C

0.37 V

D

0.30 V

Option C is Correct

Magnitude of Induced EMF in the Circuit

  • Consider a circuit in which a conducting rod of length  \(\ell\) is placed over two rail which are clamped as shown in figure.
  • The region contains uniform magnetic field \(\vec B\) and the rod is moving with velocity \(v\).
  • Since, the rod is moving with velocity \(v\), therefore all the charges moving in rod also move with velocity \(v\).

  • Due to the movement of rod, the positive charge will experience force in upward direction while negative charge will experience force in downward direction.
  • Due to the force experienced by charges, the charges get accumulated, as shown in figure.
  • The accumulation of charges creates an electric field whose direction is along (–y) axis. Due to this field, the electric force also generates.
  • The charge carriers will continue to move until the electric force \(F_E=qE\) exactly balance the magnetic force, \(F_B=qvB\)
  • The balance condition is

\(F_B=F_E\)

\(qvB=qE\)

\(E=vB\)

 

  • The potential difference across the ends of rod :

\(\Delta V=E\ell\)

\(\because\,E=vB\)

\(\therefore\,\Delta V=Bv\ell\)

  • Magnetic flux :

\(\phi=B\ell x\)

  • Emf in the circuit :

\(\dfrac{d\phi}{dt}=B\ell\dfrac{dx}{dt}\)

(\(\dfrac{dx}{dt}=v\), where \(v\) is velocity of rod)

\(\dfrac{d\phi}{dt}=B\ell v \)

\(\because\,\,Emf=\dfrac{-d\phi}{dt}\)

\(\therefore\)  The induced emf is 

\(\mathcal{E}=-B\ell v\)

Illustration Questions

In the shown figure, the rod of length \(\ell=1\,m\) is moving with velocity \(v=2\,m/s\) in the magnetic field, \(B=1\,T\). Find the magnitude of induced emf in the circuit. 

A 2 V

B 3 V

C 4 V

D 5 V

×

Length of rod, \(\ell=1\,m\)

Velocity of rod, \(v=2\,m/s\)

Magnetic field, \(B=1\,T\)

Magnitude of flux is given by

\(|\mathcal{E}|=B\ell v\)

\(|\mathcal{E}|=(1)(1)(2)\)

\(|\mathcal{E}|=2\,V\)

In the shown figure, the rod of length \(\ell=1\,m\) is moving with velocity \(v=2\,m/s\) in the magnetic field, \(B=1\,T\). Find the magnitude of induced emf in the circuit. 

image
A

2 V

.

B

3 V

C

4 V

D

5 V

Option A is Correct

Illustration Questions

A triangular coil of wire having \(N=2\) turns of side \(s=1\,m\) each, is rotating at constant frequency, \(f=60\,Hz\) in uniform magnetic field, \(B=0.1\,T\). Find the maximum induced current in the coil. Given : \(R=9.42\,\Omega\) 

A \(40\,A\)

B \(2\sqrt3\,A\)

C \(28.32\,A\)

D \(41\,A\)

×

Number of turns, \(N=2\)

Length of each side, \(s=1\,m\)

Frequency, \(f=60\,Hz\)

Magnetic field, \(B=0.1\,T\)

Resistance, \(R=9.42\,\Omega\)

maximum induced emf is given by,

\(\mathcal{E}_{max}=NAB\,\omega\)

\(=2×\dfrac{1}{2}×1×\dfrac{\sqrt3(1)}{2}×(0.1)×2\pi×60\)

\(=(18.84)\sqrt3\,V\)

Maximum current is given by,\(I_{max}=\dfrac{\mathcal{E}_{max}}{R}\)

\(I_{max}=\dfrac{18.84{\sqrt3}}{9.42}\)

\(I_{max}=2\sqrt3\,A\)

A triangular coil of wire having \(N=2\) turns of side \(s=1\,m\) each, is rotating at constant frequency, \(f=60\,Hz\) in uniform magnetic field, \(B=0.1\,T\). Find the maximum induced current in the coil. Given : \(R=9.42\,\Omega\) 

image
A

\(40\,A\)

.

B

\(2\sqrt3\,A\)

C

\(28.32\,A\)

D

\(41\,A\)

Option B is Correct

Magnetic Flux and Average E.M.F.

  • Magnetic Flux is given by 

                 \(\phi _m = \vec B \cdot \vec A\)

  • Rate of change of  flux 

             \(\left|\dfrac{d}{dt}\phi_m \right| = \left|\vec B \cdot \dfrac{d \vec A}{dt} + \vec A \cdot \dfrac{d \vec B}{dt}\right|\)

             \(\mathcal{E} = \left| \vec B \cdot \dfrac{d\,\vec A}{dt} + \vec A \cdot \dfrac{d \vec B}{dt} \right| \)

  • Due to change in flux, emf is induced in the coil.

Factors which affect Induced EMF 

i) When area vector  \(\vec A\)  is constant and magnetic field \(\vec B\)  varies 

     \(\mathcal{E} =\vec A \cdot \dfrac{d\vec B}{dt}\)

 \(\mathcal{E}_{avg} =\vec A \cdot \dfrac{\Delta\vec B}{\Delta t}\)

ii)  When magnetic field is constant and area vector varies

   \(\mathcal{E}_{avg}=\vec B \cdot \dfrac{\Delta\vec A}{\Delta t}\)

Illustration Questions

If flux in coil is \(\phi=2\,t+3\), then determine the magnitude of induced emf due to the flux in the coil. 

A +3 V

B +5 V

C +2 V

D +10 V

×

Flux in coil, \(\phi=2\,t+3\)

Rate of change of flux with time,

\(\dfrac{d\phi}{dt}=\dfrac{d}{dt}(2t+3)\)

\(\dfrac{d\phi}{dt}=2\)

\(\mathcal{E}=-\dfrac{d\phi}{dt}\)

magnitude od induced emf \(|\mathcal{E}|=|-\dfrac{d\phi}{dt}|\)

 

 

emf due to the flux,

\(\mathcal{E} =+2\,V\)

If flux in coil is \(\phi=2\,t+3\), then determine the magnitude of induced emf due to the flux in the coil. 

A

+3 V

.

B

+5 V

C

+2 V

D

+10 V

Option C is Correct

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