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.
\(\phi(t)=f(t)\) is a equation of straight line.
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
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 -
Case I :
When loop is entering in magnetic field from 0 to b width of loop :
\(\mathcal{E}=-B\ell v\)
Case II :
When loop fully enters in region of magnetic field :
\(\phi=0\)
\(\Rightarrow\,\,\dfrac{d\phi}{dt}=0\)
\(\Rightarrow\,\,\mathcal{E}=0\)
Case III :
When the loop fully left the region of \(\vec B\) :
Hence, \(\mathcal{E}=\dfrac{-d\phi}{dt}=B\ell v\)
Emf is positive.
\(\phi(t)\) with time :
\(\mathcal{E}\) with time :
If \(\theta\) is the angle between \(\vec A\) and \(\vec B\) then
\(\phi _m=BA\,cos\theta\)
\(\because\,\theta=\omega t\)
\(\therefore\,\phi _m=BA\,cos\,\omega t\)
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.
\(\mathcal{E}_{max}=NBA\,\omega \)
\(\phi _m = \vec B \cdot \vec A\)
\(\left|\dfrac{d}{dt}\phi_m \right| = \left|\vec B \cdot \dfrac{d \vec A}{dt} + \vec A \cdot \dfrac{d \vec B}{dt}\right|\)
\(\mathcal{E} = \left| \vec B \cdot \dfrac{d\,\vec A}{dt} + \vec A \cdot \dfrac{d \vec B}{dt} \right| \)
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
\(\mathcal{E}_{avg}=\vec B \cdot \dfrac{\Delta\vec A}{\Delta t}\)
A +3 V
B +5 V
C +2 V
D +10 V
\(F_B=F_E\)
\(qvB=qE\)
\(E=vB\)
\(\Delta V=E\ell\)
\(\because\,E=vB\)
\(\therefore\,\Delta V=Bv\ell\)
\(\phi=B\ell x\)
\(\dfrac{d\phi}{dt}=B\ell\dfrac{dx}{dt}\)
( \(\dfrac{dx}{dt}=v\), where \(v\) is velocity of rod)
\(\dfrac{d\phi}{dt}=B\ell v \)
\(\because\,\,emf=\dfrac{-d\phi}{dt}\)
\(\therefore\) The induced emf is,
\(\mathcal{E}=-B\ell v\)
\(\phi _m=\vec B.\,\vec A\)
\(\phi _m=BA\,cos\theta\)
\(\because\,\theta=[f(t)]\)
\(\therefore\,\phi_m=BA \,cos[f(t)]\)
\(|\mathcal{E}|=\dfrac{d\phi_m}{dt}\)
\(|\mathcal{E}|=\dfrac{d}{dt}\{BA \,cosf(t)\}\)
\(|\mathcal{E}|=BA\,\dfrac{d}{dt}\{cosf(t)\}\)
A 0.948 V
B 0.546 V
C 0.847 V
D 0.666 V
\(F_B=F_E\)
\(qvB=qE\)
\(E=vB\)
\(\Delta V=E\ell\)
\(\because\,E=vB\)
\(\therefore\,\Delta V=Bv\ell\)
\(\phi=B\ell x\)
\(\dfrac{d\phi}{dt}=B\ell\dfrac{dx}{dt}\)
(\(\dfrac{dx}{dt}=v\), where \(v\) is velocity of rod)
\(\dfrac{d\phi}{dt}=B\ell v \)
\(\because\,\,Emf=\dfrac{-d\phi}{dt}\)
\(\therefore\) The induced emf is
\(\mathcal{E}=-B\ell v\)