Informative line

Induced Electric Field

Learn direction of induced current and electric field and practice to determine the direction of induced electric field lines at point. Practice to find the magnitude of induced electric field at a distance from center inside and outside the solenoid.

Concept of Induced Electric Field

  • When a loop or coil is placed in varying magnetic field, the current is induced into the coil.
  • Direction of Induced Current

    Direction of induced current is given by Lenz's law.

  • This current is produced by an electric field. The electric field makes charges to move.
  • The electric field is caused by changing magnetic field and is called an induced electric field.
  • The space in which the magnetic field is changing, the pinwheel pattern of electric field will generate as shown in figure.  

  

Direction of Induced Electric Field 

The direction of induced electric field should be such that it drives the current in a loop in the direction which opposes the change in magnetic field.

Illustration Questions

A solenoid is placed in increasing magnetic field such that the top of solenoid lies in x - y plane, as shown in figure. Determine the direction of induced electric field at point P.

A Along  \(-\hat i\)

B Along + \(\hat j\)

C Along + \(\hat i\)

D Along (– \(\hat j\))

×

The direction of induced current in the loop is determined by using Lenz's law.

The direction of induced current will be anti-clockwise to oppose the change in magnetic field.

image

The direction of induced electric field is same as the direction of induced current in the loop.

Thus, the direction of induced electric field is along +\(\hat j\) at point P. 

A solenoid is placed in increasing magnetic field such that the top of solenoid lies in x - y plane, as shown in figure. Determine the direction of induced electric field at point P.

image
A

Along  \(-\hat i\)

.

B

Along + \(\hat j\)

C

Along + \(\hat i\)

D

Along (– \(\hat j\))

Option B is Correct

Properties of Induced Electric Field Lines 

  • Induced electric field lines are not created by source charges.
  • These are created by changing magnetic field.
  • These lines  must form closed loops without any source charge present in loop.
  • The electric field vector \((\vec{E})\) is tangent to the electric field line at each point and its direction is given by the direction of field line. 

  • Electric field lines make no sharp turn because if they do so then tangent at that point will be undefined and hence electric field vector is also undefined.

This type of electric field lines is not possible.  

  • Electric field lines do not intersect each other because then electric field vector will have two directions at that point.

  • Electric field lines are non - conservative in nature i.e. , work done along path (circle) depends upon distance. 

          \(W= E_{ind} × (distance)\) 

          \(W= E_{ind} × (2\,\pi\,R)\)

Illustration Questions

Which of the following is not the property of induced electric field lines?

A Tangent at any point on electric field line gives the direction of electric field vector

B Electric field lines never cross each other

C Electric field lines do not form closed loops

D Electric field lines make no sharp turn

×

Options (A), (B) and (D) are the properties of induced electric field lines. 

Option (C) is incorrect because electric field lines form closed loops. 

Which of the following is not the property of induced electric field lines?

A

Tangent at any point on electric field line gives the direction of electric field vector

.

B

Electric field lines never cross each other

C

Electric field lines do not form closed loops

D

Electric field lines make no sharp turn

Option C is Correct

Induced Electric Field Inside a Solenoid

  • Consider a long solenoid of radius R having n turns of wire per unit length and magnetic field   \(\vec{B} = B_0t. \)
  • We will calculate magnitude of induced electric field at a distance ' r ' from its center.
  • The magnetic field is changing. Thus, induced electric field will be created.
  • Work done by electric field in moving a test charge q is given by 

           \(W_1=q\;\mathcal{E}\) 

             where \(\mathcal{E}\) is the induced emf

  • Work done by electric field to move charge on the periphery of loop 

             \(W_2=q\;E(2\pi \;r)\) 

            where  \(2 \,\pi r\) is the circumference of loop

As \(E_{ind}\) is non conservative , so work is non zero and depends upon distance.

  • Both work done are equal thus, 

                            \(q\mathcal{E} = q E\;(2 \pi r)\)

                            \(\mathcal{E} = E. 2 \pi r \;\;\;\;— (1)\)

  • Since,       \(\mathcal{E}= \dfrac {-d\phi}{dt}\;\;\;\;\;\;— (2)\)
  • From equation (1) and (2)

                   \(E. 2 \pi r = \dfrac {-d\phi}{dt}\)

           \(\because\)     \(\phi = B.A \)  and  \( A =\pi r^2\)

            \(\therefore\)     \(E. 2 \pi r = - \pi r^2 \dfrac {dB}{dt}\)

                   \(E = \dfrac {-r}{2}\dfrac{dB}{dt}\)     

  • The EMF for any Closed Path

              \(\oint \vec{E}.\;d\vec{S}\) = Induced EMF

           According to Faraday's law —

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

                Thus,      \(\oint \vec{E}.\;d\vec{S}\;=\;\dfrac {-d\phi}{dt}\)

  • To calculate induced electric field, we take integration loop of radius r.
  • When r < R (radius of solenoid):

Then flux through integration loop is 

   \(\dfrac {-d\phi}{dt}\;=\;\dfrac{-d}{dt}\;\;(B.\pi r ^2)\)

   \(\dfrac {-d\phi}{dt}\;=\; -\pi r^2\dfrac {d}{dt}(B_0t)\)

    \(\dfrac {-d\phi}{dt}= - B_0\;\pi r^2\)

The top view of solenoid is shown in figure.

  • As,    \(\oint \vec{E}.\;d\vec{S}\;=\;\dfrac {-d\phi}{dt}\)

\(\oint \vec{E}.\;d\vec{S} = -B_0\pi r^2\)

\(E. 2 \pi r= - B_0 \pi r ^2\)

\(E_{inside } = \dfrac {-B_0\;r}{2}\)

\(|E_{inside }| = \dfrac {B_0\;r}{2}\)

  • The magnitude of electric field inside the solenoid varies linearly with 'r' when magnetic field B is a linear function of time.

Illustration Questions

A solenoid of radius R = 5m has increasing magnetic field  \(B = 3\mu_0t\). Find the magnitude of induced electric field at distance r = 2m from center inside the solenoid.  

A  \(3\,\mu _0\;V/m \)  

B  \(5\,\mu _0V/m \) 

C  \(10\,\mu _0V/m \)

D  \(7\,\mu _0V/m \) 

×

Magnetic field of solenoid, \( B = 3\mu _0 t\)

 \(\dfrac {dB}{dt} \;=\;3\mu_0\)

Since, \(\oint \vec{E}.\;d\vec{S}\;=\;\dfrac {-d\phi}{dt}\)

\( \vec{E}\oint d\vec{S}\;=\;-A\dfrac {dB}{dt}\)

 \(E(2 \pi r )= - \pi r^2 \;\;3 \mu_0\)

\(E = \dfrac{-3 \mu_0 r}{2}\)

\(\because\)  \( r = 2 m\)

\(\therefore \;\;|E| = \dfrac {3}{2}\mu_0 × 2 \)

\(|E| = 3 \,\mu_0\,V/m\)

A solenoid of radius R = 5m has increasing magnetic field  \(B = 3\mu_0t\). Find the magnitude of induced electric field at distance r = 2m from center inside the solenoid.  

A

 \(3\,\mu _0\;V/m \)  

.

B

 \(5\,\mu _0V/m \) 

C

 \(10\,\mu _0V/m \)

D

 \(7\,\mu _0V/m \) 

Option A is Correct

Induced Electric Field Outside a Solenoid

  • Consider a long solenoid of radius R having n turns of wire per unit length and magnetic field   \(\vec{B} = B_0t. \)
  • We will calculate magnitude of induced electric field at a distance ' r ' from its center.
  • The magnetic field is changing. Thus, induced electric field will be created.

Work done by electric field in moving a test charge q is given by 

           \(W_1=q\;\mathcal{E}\) 

             where \(\mathcal{E}\) is the induced emf

Work done by electric field to move charge on the periphery of loop 

             \(W_2=q\;E(2\pi \;r)\) 

            where  \(2 \,\pi r\) is the circumference of loop

As \(E_{ind}\) is non conservative , so work is non zero and depends upon distance.

  • Both work done are equal thus, 

                            \(q\mathcal{E} = q E\;(2 \pi r)\)

                            \(\mathcal{E} = E. 2 \pi r \;\;\;\;— (1)\)

  • Since,       \(\mathcal{E}= \dfrac {-d\phi}{dt}\;\;\;\;\;\;— (2)\)
  • From equation (1) and (2)

                   \(E. 2 \pi r = \dfrac {-d\phi}{dt}\)

           \(\because\)     \(\phi = B.A \)  and  \( A =\pi r^2\)

            \(\therefore\)     \(E. 2 \pi r = - \pi r^2 \dfrac {dB}{dt}\)

 

                    \(E = \dfrac {-r}{2}\dfrac{dB}{dt}\)     

  • The EMF for any Closed Path

              \(\oint \vec{E}.\;d\vec{S}\) = Induced EMF

           According to Faraday's law —

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

                Thus,      \(\oint \vec{E}.\;d\vec{S}\;=\;\dfrac {-d\phi}{dt}\)

  • To calculate induced electric field, we take integration loop of radius r.
  • When r > R (radius of solenoid):

    Then, flux through integration loop is 

    \(\dfrac {-d\phi}{dt} = \dfrac {-d}{dt} (B \pi R^2)\)

    \(\dfrac {-d\phi}{dt} = - \pi R^2 \dfrac {d}{dt}(B_0t)\)

    \(-\dfrac {d\phi}{dt} = - B_0\pi R^2\)  

The top view of solenoid is shown in figure.

  • As, \(\oint\vec{E}. d \vec{S} = \dfrac {-d\phi}{dt}\)

\(\oint\vec{E}. d \vec{S} = - B_0\pi R ^ 2 \)

\(E( 2 \pi r) = - B_0\pi R^2\)

\(E _{outside }= \dfrac{-B_0\pi R^2}{2 \pi r}\)

\(|E_{outside}| = \dfrac{B_0R^2}{2r}\)  

The magnitude of electric field outside the solenoid falls of as 1/r, when magnetic field \(\vec{B}\) is a linear function of time.

Illustration Questions

A solenoid of radius R = 5m has increasing magnetic field  \(B = 3\,\mu_0t\). Find the magnitude of induced electric field at a distance r = 10 m from center outside the solenoid.

A  \(5\,\mu _0V /m\)  

B \(\dfrac {15}{4}\) \(\mu _0 \,V/m\)  

C   \(7\,\mu _0 V/m\) 

D  \(8\,\mu _0 V/m\)  

×

Magnetic field of solenoid,  \( B = 3\mu _0 t\)

 \(\dfrac{dB}{dt} \;=\; 3 \mu _0\) 

Since, \(\oint\vec{E}. d \vec{S}\;=\;\dfrac{-d\phi}{dt}\)

\(\vec{E}\oint d \vec{S} = -A \dfrac {dB}{dt}\)

 \(E( 2\pi r)\) = – \(\pi R^2\;3 \mu _0\)

\(E = \dfrac{-3 \mu_0R^2}{2 r }\)

\(\because \) \( r = 10 m , R = 5 m\)

\(\therefore \;\;|E| = \dfrac{3 \mu_0(5)^2}{2 × 10 }\)

\(|E| \;\;= \dfrac{15}{4}\;\mu_0\,V/m\)

A solenoid of radius R = 5m has increasing magnetic field  \(B = 3\,\mu_0t\). Find the magnitude of induced electric field at a distance r = 10 m from center outside the solenoid.

A

 \(5\,\mu _0V /m\)  

.

B

\(\dfrac {15}{4}\) \(\mu _0 \,V/m\)  

C

  \(7\,\mu _0 V/m\) 

D

 \(8\,\mu _0 V/m\)  

Option B is Correct

Illustration Questions

A long solenoid of radius R = 5m has n = 1000 turns of wire per unit length carries a time varying current  \( I = 3t + 2\) . Determine the magnitude of induced electric field inside the solenoid at a distance r = 2m from center.    

A \(12 \)\(\pi × 10 ^{-4}\) \(V/m\)

B \(8\)\(\pi × 10 ^{-4}\) \(V/m\)

C \(18\; \pi × 10 ^{-4}\) \(V/m\)

D \(10\) \(\pi × 10 ^{-4}\) \(V/m\)

×

Magnetic field of solenoid, \(B = \mu_0n\) \(I\) 

 \(B =\mu_0\) \((1000) ( 3t + 2 )\)

\(\dfrac {dB}{dt}\)\( = 1000\)\(\mu_0\)\( (3)\)

Since, \(\oint \vec{E}.d\vec {S} = \dfrac {-d\phi}{dt}\)

\( \vec{E} \oint d\vec {S} = \dfrac {-A dB}{dt}\)

\(E(2\) \(\pi r)\) = – \(\pi r^2\) \((3) (1000)\) \(\mu_0\)

\(E =\dfrac {-(3000)\mu_0\, r}{2}\)

    \(\because\;\;\;r=2m\)

\(\therefore\;\;\)\(|E| = \dfrac {(3000)(4 \pi × 10 ^ {-7})(2)}{2}\)

\(\)  \(|E| = 12 \pi × 10 ^{-4}\) \(V/m\)

A long solenoid of radius R = 5m has n = 1000 turns of wire per unit length carries a time varying current  \( I = 3t + 2\) . Determine the magnitude of induced electric field inside the solenoid at a distance r = 2m from center.    

A

\(12 \)\(\pi × 10 ^{-4}\) \(V/m\)

.

B

\(8\)\(\pi × 10 ^{-4}\) \(V/m\)

C

\(18\; \pi × 10 ^{-4}\) \(V/m\)

D

\(10\) \(\pi × 10 ^{-4}\) \(V/m\)

Option A is Correct

Direction of Force due to Induced Electric Field

  • Consider an electron that is placed in a varying magnetic field.
  • As magnetic field is varying, an electric field is induced, as shown in figure.

  • The direction of force due to induced electric field is opposite to the direction of induced electric field on electron(i.e., negative charge).

Illustration Questions

In a region of decreasing magnetic field, as shown in figure, determine the direction of force due to induced electric field on an electron placed in such system.

A Anti-clockwise 

B Clockwise 

C No force 

D None of these

×

Since, magnetic field is decreasing, thus by Lenz's law, the direction of induced electric field is clockwise. 

The direction of force due to induced electric field is opposite to the direction of induced electric field on electron (i.e. negative charge).

Hence, direction of force due to induced electric field is anti-clockwise.

In a region of decreasing magnetic field, as shown in figure, determine the direction of force due to induced electric field on an electron placed in such system.

image
A

Anti-clockwise 

.

B

Clockwise 

C

No force 

D

None of these

Option A is Correct

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