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.
Direction of induced current is given by Lenz's law.
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.
A Along \(-\hat i\)
B Along + \(\hat j\)
C Along + \(\hat i\)
D Along (– \(\hat j\))
This type of electric field lines is not possible.
\(W= E_{ind} × (distance)\)
\(W= E_{ind} × (2\,\pi\,R)\)
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
A \(12 \)\(\pi × 10 ^{-4}\) \(V/m\)
B \(8\)\(\pi × 10 ^{-4}\) \(V/m\)
C \(18\; \pi × 10 ^{-4}\) \(V/m\)
D \(10\) \(\pi × 10 ^{-4}\) \(V/m\)
A Anti-clockwise
B Clockwise
C No force
D None of these
\(W_1=q\;\mathcal{E}\)
where \(\mathcal{E}\) is the induced emf
\(W_2=q\;E(2\pi \;r)\)
where \(2 \,\pi r\) is the circumference of loop
As \(E_{ind}\) is non conservative , so work is non zero and depends upon distance.
\(q\mathcal{E} = q E\;(2 \pi r)\)
\(\mathcal{E} = E. 2 \pi r \;\;\;\;— (1)\)
\(E. 2 \pi r = \dfrac {-d\phi}{dt}\)
\(\because\) \(\phi = B.A \) and \( A =\pi r^2\)
\(\therefore\) \(E. 2 \pi r = - \pi r^2 \dfrac {dB}{dt}\)
\(E = \dfrac {-r}{2}\dfrac{dB}{dt}\)
\(\oint \vec{E}.\;d\vec{S}\) = Induced EMF
According to Faraday's law —
Induced emf = \(\dfrac {-d\phi}{dt}\)
Thus, \(\oint \vec{E}.\;d\vec{S}\;=\;\dfrac {-d\phi}{dt}\)
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.
\(\oint \vec{E}.\;d\vec{S} = -B_0\pi r^2\)
\(E. 2 \pi r= - B_0 \pi r ^2\)
\(E_{inside } = \dfrac {-B_0\;r}{2}\)
\(|E_{inside }| = \dfrac {B_0\;r}{2}\)
A \(3\,\mu _0\;V/m \)
B \(5\,\mu _0V/m \)
C \(10\,\mu _0V/m \)
D \(7\,\mu _0V/m \)
\(W_1=q\;\mathcal{E}\)
where \(\mathcal{E}\) is the induced emf
\(W_2=q\;E(2\pi \;r)\)
where \(2 \,\pi r\) is the circumference of loop
As \(E_{ind}\) is non conservative , so work is non zero and depends upon distance.
\(q\mathcal{E} = q E\;(2 \pi r)\)
\(\mathcal{E} = E. 2 \pi r \;\;\;\;— (1)\)
\(E. 2 \pi r = \dfrac {-d\phi}{dt}\)
\(\because\) \(\phi = B.A \) and \( A =\pi r^2\)
\(\therefore\) \(E. 2 \pi r = - \pi r^2 \dfrac {dB}{dt}\)
\(E = \dfrac {-r}{2}\dfrac{dB}{dt}\)
\(\oint \vec{E}.\;d\vec{S}\) = Induced EMF
According to Faraday's law —
Induced emf = \(\dfrac {-d\phi}{dt}\)
Thus, \(\oint \vec{E}.\;d\vec{S}\;=\;\dfrac {-d\phi}{dt}\)
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.
\(\oint\vec{E}. d \vec{S} = - B_0\pi R ^ 2 \)
\(E( 2 \pi r) = - B_0\pi R^2\)
\(E _{outside }= \dfrac{-B_0\pi R^2}{2 \pi r}\)
\(|E_{outside}| = \dfrac{B_0R^2}{2r}\)
The magnitude of electric field outside the solenoid falls of as 1/r, when magnetic field \(\vec{B}\) is a linear function of time.
A \(5\,\mu _0V /m\)
B \(\dfrac {15}{4}\) \(\mu _0 \,V/m\)
C \(7\,\mu _0 V/m\)
D \(8\,\mu _0 V/m\)