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Average Value Of Induced Emf

Learn formula of average induced EMF and magnetic flux, calculate average induced EMF inside the solenoid. Practice to find the magnitude of average induced EMF in that period of time.

Induced EMF

  • Consider a conducting coil by which galvanometer is connected.
  • A bar magnet is placed, as shown in figure.

 

 

Different cases can be applied

Case 1   When bar magnet is stationary

          The magnetic flux linked with coil will be constant as the number of field lines passing through coil is constant.

Case 2   When bar magnet is moved closer to coil 

The magnetic flux linked with coil will increase as the number of field lines passing through coil is increasing.

Case 3   When bar magnet is moved away from coil 

The magnetic flux linked with coil will decrease as the number of field lines passing through coil is decreasing.

  • By these three cases, it is clear that rate of change of flux sets up a current in the coil.
  • This current is known as induced current and it is produced by induced emf.

  • The induced emf is directly proportional to the rate of change of flux through the loop.

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

                   \(\mathcal{E} = \dfrac{- d }{dt}(BA\,cos\theta)\)     [\(\because\) \(\phi_m = BA \,cos\theta\)]

  • Emf induced by a coil consisting of N loops with the same area is given as

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

Condition of Induced EMF 

  • Magnetic field  \(\vec B\) varies with time.
  • Area enclosed by loop varies with time.
  • The angle \(\theta\) between magnetic filed \(\vec B\) and area vector \(\vec A\) varies with time.

Illustration Questions

In which of the following option, induced emf is more?

A

B

C

D

×

Since, induced emf is directly proportional to rate of change of flux.

image

The number of field lines passing through the loop is more in A, as both are moving  towards each other. 

So, flux is more in option A. 

Hence, induced emf is more in A.

So, option A is correct.

In which of the following option, induced emf is more?

A image
B image
C image
D image

Option A is Correct

Magnetic Flux and Average EMF

  • 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

A square loop of side a= 2 m is placed in a uniform magnetic field directed perpendicular to the plane of the loop. If the field changes linearly from 0 to 2 Tesla in \(\Delta t = 2 \,sec\), calculate the magnitude of induced emf in the loop.  

A 2 V 

B 4 V 

C 5 V

D 8 V

×

Average induced  emf is given as rate of change of flux

\(|\mathcal E_{avg}| = \dfrac{\Delta \phi }{\Delta t}\)

where, \(\phi = flux= BA \,cos\theta\) 

             t = time 

            B = magnetic field

            A = Area

Flux \(\phi = BA \,cos\theta\)

     where B = magnetic field 

                  A = Area

So, induced emf

\(|\mathcal E_{avg}| = A \dfrac{\Delta B }{\Delta t} = A \dfrac{(B_f - B_i)}{\Delta t}\)

 

Given:

\(a=2\,m\ ,B_f=2\ T\ ,B_i=0\ T\ ,\Delta t=2 \, sec\)

Area  A = a2 = (2 m)2 = 4 m2 

\(|\mathcal E_{avg}| = 4 × \dfrac{(2–0)}{2} = 4\, V\)

A square loop of side a= 2 m is placed in a uniform magnetic field directed perpendicular to the plane of the loop. If the field changes linearly from 0 to 2 Tesla in \(\Delta t = 2 \,sec\), calculate the magnitude of induced emf in the loop.  

A

2 V 

.

B

4 V 

C

5 V

D

8 V

Option B is Correct

Effect of Flipping of Coil in Magnetic Field on Magnetic Flux

  • Consider a coil of area A which is placed in uniform magnetic field.
  • The coil is flipped by some angle.

Effect of flipping 

          The magnetic field and area remains constant but due to flipping , the magnetic flux through the coil changes as the angle between the area vector and magnetic field changes.

At initial position

Magnetic flux is  

               \((\phi_m)_i = BA\,cos 0^\circ\)    

               \((\phi_m)_i = BA\)

          

                       

  • Let it be flipped by 180° .

At final position 

Magnetic flux is

               \((\phi_m)_f = BA\,cos 180°\)

              \((\phi_m)_f = -B\,A\)

Change in flux 

             \(\Delta \phi_m = |(\phi_m)_f-(\phi_m)_i|\)

             \(\Delta \phi_m = |-BA-BA|\)

             \(\Delta \phi_m = 2\,BA\)

  • Thus, change in flux = \(2\ BA\)
  • Let this change in flux takes place in \(\Delta t\) sec.

            Then average induced emf,

            \(\mathcal E_{avg} = \left|\dfrac{\Delta \phi}{\Delta t}\right|\)  

            \(\mathcal E_{avg} = \left|\dfrac{2BA}{\Delta t}\right|\)

Note: 

  • We can assume the direction of induced current in any direction, as it comes with its sign while calculating.
  • Direction of area vector is given by right hand thumb rule.
  • It states that using your right hand, point your thumb in  the direction of current, then the four curled fingers represents the direction of magnetic field.

Illustration Questions

A rectangular coil of area A = 2 m2 is placed vertically in uniform magnetic field B = 1 T. If the coil is flipped by \(\theta = 180°\) in \(\Delta t = 2 \,sec\), find the magnitude of average induced emf in that period of time.  

A 2 V

B 3 V

C 5 V

D 7 V

×

Flux linked with coil before flipping \(\phi_i =BA \,cos\theta\)

Since, \(\vec B \) and  \(\vec A \)  both are parallel ,i.e., \(\theta = 0°\)

\(\phi_i =BA \,cos\ 0 °\)

\(\phi_i =BA \)

\(\phi_i =1×2 = 2\,Wb\)

image

Flux linked with coil after flipping \(\phi _f =BA \, cos \theta\)

Since,the angle between  \(\vec B \) and  \(\vec A \)  is 180°.

\(\phi _f =BA \, cos 180^0 = -BA\)

 \(\phi _f =-1× 2 = -2\ Wb\)

image

Thus change is flux

\(\Delta \phi = |\phi_f-\phi_i| = |-2-2| = 4\ Wb\)

So, Average induced e.m.f.

 \(|\mathcal E_{avg}| = \left|\dfrac{\Delta\phi}{\Delta t}\right| = \dfrac{4}{2}\)

\(|\mathcal E_{avg}| = 2 \,V\)

A rectangular coil of area A = 2 m2 is placed vertically in uniform magnetic field B = 1 T. If the coil is flipped by \(\theta = 180°\) in \(\Delta t = 2 \,sec\), find the magnitude of average induced emf in that period of time.  

A

2 V

.

B

3 V

C

5 V

D

7 V

Option A is Correct

Magnitude of Induced EMF when Area of Loop Changes Linearly

  • 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 

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

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

ii)  When magnetic field is constant and area vector varies

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

  • Consider a square loop placed in a uniform magnetic field directed perpendicular to the plane of loop as shown in figure.

  • The loop is moving  with velocity \(v\) as shown in figure.
  • When loop moves towards right, area increases and thus in turn the induced emf also increases.
  • When loop moves towards left, area decreases and thus in turn the induced emf decreases.

Average induced emf is given as 

\(\left|\mathcal E_{avg}\right| = \dfrac{\Delta\phi}{\Delta t}\)

\(\left|\mathcal E_{avg}\right| = \dfrac{\Delta(BA)}{\Delta t}\)

\(\left|\mathcal E_{avg}\right| = B\left(\dfrac{\Delta A}{\Delta t}\right)\)

\(\left|\mathcal E_{avg}\right| = \dfrac{B(A_f - A_i)}{\Delta t}\)

 

 

Illustration Questions

A square loop is placed in a uniform magnetic filed of B = 2 T, directed perpendicular to the plane of the loop whose area changes linearly from \(A_i = 0\,m^2\)  to \(A_f = 4\,m^2\) in \(\Delta t = 2 \,sec. \) Calculate the magnitude of induced emf in loop.

A 3 V

B 6 V

C 5 V

D 4 V

×

Average induced emf is given as 

\(\left|\mathcal E_{avg}\right| = \dfrac{\Delta\phi}{\Delta t}\)

\(\left|\mathcal E_{avg}\right| = \dfrac{\Delta(BA)}{\Delta t}\)

\(\left|\mathcal E_{avg}\right| = B\left(\dfrac{\Delta A}{\Delta t}\right)\)

\(\left|\mathcal E_{avg}\right| = \dfrac{B(A_f - A_i)}{\Delta t}\)

Given : \(B=2\ T\ ,A_i=0\ ,A_f=4\ m^2\ ,\Delta t=2\ sec\)

\(|\mathcal E_{avg}| = \dfrac{2(4-0)}{2}\)

\(|\mathcal E_{avg}| = 4\ V\)

A square loop is placed in a uniform magnetic filed of B = 2 T, directed perpendicular to the plane of the loop whose area changes linearly from \(A_i = 0\,m^2\)  to \(A_f = 4\,m^2\) in \(\Delta t = 2 \,sec. \) Calculate the magnitude of induced emf in loop.

A

3 V

.

B

6 V

C

5 V

D

4 V

Option D is Correct

Average Induced EMF Inside The Solenoid

  • Consider a solenoid having n turns.
  • The magnetic filed of solenoid is given by 

                  \(B = \mu_0\,nI\)

 

  • Average change in magnetic field is given as 

             \(\dfrac{\Delta B}{\Delta t} = \mu_0 \,n \dfrac{\Delta I}{\Delta t}\) 

  • From equation it is clear that if we vary current, magnetic field also vary.

Illustration Questions

A loop of wire, having diameter d = 2 cm is placed at the center of the solenoid. If the solenoid has n = 1000 turns/meter and the current varies linearly from \(I_i = 0\ A\) to \(I_f = 2\ A\) in \(\Delta t = 2\ sec\). Calculate the magnitude of e.m.f. induced in the loop.   

A \(4 \pi^2 × 10^{-8 }\ V\)

B \(2 \pi^2 × 10^{-6 }\ V\)

C \(8 \pi^2 × 10^{-3 }\ V\)

D \(2 \pi^2 × 10^{-4 }\ V\)

×

Magnetic field for solenoid is given as 

\(B = \mu_0\,nI\)

Rate change of magnetic field 

\(\dfrac{\Delta B}{\Delta t} = \mu_0\,n \dfrac{\Delta I}{\Delta t}\)

Average induced e.m.f \((\mathcal{E})\) is given as rate of change of flux

\(\mathcal{E}_{avg} = \dfrac{\Delta\phi}{\Delta t}\)

\(\mathcal{E} = A \cdot\dfrac{\Delta B}{\Delta t}\)   \([\phi = AB]\)

\(\mathcal{E} = A \mu_0 \,n \dfrac{\Delta\ I}{\Delta t}\)

where, A = Area

             n = no of turns

              I = Current 

              t = time

Given :  \(n=1000\ turns\ ,I_i= 0\ A\ ,I_f = 2\ A , \, \Delta t = 2\ sec \)

\(\mathcal{E} = \pi × (1×10^{-2})^2 × 4\pi × 10^{-7} × 1000 × \dfrac{(2-0)}{2}\)

\(\mathcal{E} = 4\pi^2 × 10^{-8}\ V\)

A loop of wire, having diameter d = 2 cm is placed at the center of the solenoid. If the solenoid has n = 1000 turns/meter and the current varies linearly from \(I_i = 0\ A\) to \(I_f = 2\ A\) in \(\Delta t = 2\ sec\). Calculate the magnitude of e.m.f. induced in the loop.   

A

\(4 \pi^2 × 10^{-8 }\ V\)

.

B

\(2 \pi^2 × 10^{-6 }\ V\)

C

\(8 \pi^2 × 10^{-3 }\ V\)

D

\(2 \pi^2 × 10^{-4 }\ V\)

Option A is Correct

Flux Linked with One Solenoid Due to Other  

  • Consider two coaxial solenoids such that solenoid of radius \(r_1\) is placed inside a solenoid of radius \(r_2\), coaxially.

  • The number of turns in inner and outer solenoid is \(N_1 \,\,\text{and }\,\, N_2\), respectively.

Case 1:

  • In outer solenoid, current varies linearly from \(I_1 \,\,\text{to }\,\, I_2\) in time \(\Delta t\).
  • Area of inner solenoid,

 \(A= \pi\,r_1^2 × \text{number of turns}\)

\(= N_1 \,\pi\,r_1^2\)

  • The magnetic field inside the outer solenoid is given by 

\(B=\mu_0 N_2I\)

  • Then, the flux due to outer solenoid at inner solenoid

\(\phi=B.A\)

  • The rate of change of flux due to outer solenoid at inner solenoid

\(\dfrac{\Delta \phi}{\Delta t} = \dfrac{\Delta (B.A)}{\Delta t}\)

\(\dfrac{\Delta \phi}{\Delta t} = A\dfrac{\Delta B}{\Delta t}\)

\(\dfrac{\Delta \phi}{\Delta t} = N_1\,\pi\,r_1^2 . \mu_0\,N_2\,\dfrac{\Delta \,I}{\Delta t}\)

 

Case 2:

  • If, in inner solenoid, current varies linearly from \(I_1 \,\,\text{to }\,\, I_2\) in time \(\Delta t\), then area of inner solenoid, 

\(A= \pi\,r_1^2 × \text{number of turns}\)

\(A= \pi\,r_1^2× N_2\)

  • The magnetic field inside the inner solenoid is given by 

\(B= \mu_0\,N_1 \,I\)

  • Then flux due to inner solenoid at outer solenoid 

\(\phi =B.A\)

  • The rate of change of flux due to inner solenoid at outer solenoid,

\(\dfrac{\Delta \phi}{\Delta t} =\dfrac{\Delta }{\Delta t}(B.A) \)

\(\dfrac{\Delta \phi}{\Delta t} =A\left(\dfrac{\Delta B }{\Delta t}\right)\)

\(\dfrac{\Delta \phi}{\Delta t} =\pi\,r_1^2 \,N_2 \,\,\mu_0\,N_1\dfrac{\Delta I }{\Delta t}\)

 

Illustration Questions

A, \(d_i = 2\ m \) diameter solenoid is placed inside a solenoid of diameter \(d_0 = 4\ m\). Inner solenoid has  \(n_i = 1000 \,turns/length\) and outer solenoid has  \(n_0 = 2000 \,turns/length\). If current in outer solenoid varies linearly from \(I _i = 0\,A\)  to \(I _f = 2\ A\) in \(\Delta t = 2 \,sec\), calculate the value of average flux change due to outer  solenoid at inner solenoid.  

A \(0\cdot8 \pi^2 \,Wb\)

B \(0\cdot4 \pi^2 \,Wb\)

C \(0\cdot6 \pi^2 \,Wb\)

D \(0\cdot9 \pi^2 \,Wb\)

×

 \(Area\ of\ inner\ solenoid=\pi r^2\times number\ of\ turns(n_i)\) 

                                   \(= \pi (1)^2 × 1000\)

                                  \(= 1000 \pi\,m^2\)

 

 

Rate of charge of flux due to outer solenoid at inner solenoid 

\(\dfrac{\Delta \phi}{\Delta t} = A \dfrac{\Delta I}{\Delta t}\)   

\(\dfrac{\Delta \phi}{\Delta t} = A \mu_0n\dfrac{\Delta I}{\Delta t}\)  

[  \(\therefore B = \mu_0\, nI\)]

where A = Area of inner solenoid 

           n = number of turns of other solenoid 

          B = magnetic field 

           I =  Current

Given :  \(n_0 = 2000\ ,I_f = 2\ A\ ,I_i = 0\ A\ ,\Delta t = 2 \,sec\) 

 \(\dfrac {\Delta \phi}{\Delta t} = 1000\pi × 4\pi× 10^{-7}× 2000× \dfrac{2}{2}\) 

\(\dfrac{\Delta \phi }{\Delta t} = \cdot 8 \pi^2 \, Wb\)

A, \(d_i = 2\ m \) diameter solenoid is placed inside a solenoid of diameter \(d_0 = 4\ m\). Inner solenoid has  \(n_i = 1000 \,turns/length\) and outer solenoid has  \(n_0 = 2000 \,turns/length\). If current in outer solenoid varies linearly from \(I _i = 0\,A\)  to \(I _f = 2\ A\) in \(\Delta t = 2 \,sec\), calculate the value of average flux change due to outer  solenoid at inner solenoid.  

A

\(0\cdot8 \pi^2 \,Wb\)

.

B

\(0\cdot4 \pi^2 \,Wb\)

C

\(0\cdot6 \pi^2 \,Wb\)

D

\(0\cdot9 \pi^2 \,Wb\)

Option A is Correct

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