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This is the second of our many quizzes on Electromagnetic induction. Click on this to access more such competitive quizzes on Current Electricity and EMI.

Electromagnetic induction Quiz 2

physics

A coil having $N=20$ turns , length $L=2.5\,m$ and cross-sectional area of $0.5\,{{m}^{2}}$. The change in current flowing with respect time is $0.5\,A/s$ find the emf induced in the coil.$[\text{assuming permeability of free space is 1}]$ 

The rate of current flowing refers to current flowing through wire per second. 

  $\overrightarrow{B}={{\mu }_{\circ }}ni\\\Rightarrow \phi =\overrightarrow{B}A\\\Rightarrow e=\dfrac{d\phi }{dt}\\\text{Putting the value of}\,\phi\,\text{and}\, \overrightarrow{B}\,\text{in above equation we have-}\\e=\dfrac{d(\overrightarrow{B}A)}{dt}\\\Rightarrow e =\dfrac{d({{\mu }_{\circ }}niA)}{dt}={{\mu }_{\circ }}nA\dfrac{di}{dt}\\\Rightarrow e =4\pi \times {{10}^{-7}}\times \dfrac{20}{2.5}\times 0.5\times 0.5\\\therefore e=25.12\times {{10}^{-7}}\,V$

Two friends $A$ and $B$ having 2 coils of $N=5$ and $N=13$. They started playing a game in which a varying magnetic flux is supplied to both coils. A’s varying magnetic flux in coil 1 is $\phi =4{{t}^{2}}+3t+2$ and B’s varying magnetic flux in coil 2 is  $\phi =9t+4$. The relation between induced emf in both coils i.e ${{V}_{1}}$ and ${{V}_{2}}$ after 5 sec is-

The change in flux is maximum in coil one so emf generated will be more in coil one than two. 

  $e=N\dfrac{d(\phi )}{dt}\\\text{For coil 1 we have-}\\{{e}_{1}}=N\dfrac{d(4{{t}^{2}}+3t+2)}{dt}=5(8t+3)\\\text{At t=5sec}\\{{e}_{1}}=5\times (8\times 5+3)\\\Rightarrow {{e}_{1}}=215V\\\text{For coil 2}\\{{e}_{2}}=N\dfrac{d(9t+4)}{dt}\\\Rightarrow {{e}_{2}} =13\times 9\\\therefore {{e}_{2}} =117\,V$

One day Sofia went to a park where she saw a rod of length $1\,m$ falling freely from a tree of height 8 m and she observed a pattern in it by which the length of the rod remains perpendicular to the earth magnetic field (40400\,nT)  throughout its journey. Now she wants to calculate the difference of emf induced in the rod in two positions: $\\$ 1) $3\,m$ above the ground level $\\$ 2) When it just reaches the ground level.$\\$ The magnitude of difference of emf induced in above two condition will be-

The emf induced in a rod increases as the velocity increases as it approaches the ground. 

 $e=Blv\\\text{In first case rod travels 5 m from top so,}\\{{v}^{2}}={{u}^{2}}+2as\\\Rightarrow {{v}^{2}}={{0}^{2}}+2\times 10\times 5\\\Rightarrow v=10\,m/s\\\Rightarrow e=40.40\times {{10}^{-6}}\times 1\times 10=404\times {{10}^{-6}}V\\\text{In second case,}\\mgh=\dfrac{1}{2}m{{v}^{2}}\\\Rightarrow 10\times 8=\dfrac{1}{2}\times {{v}^{2}}\\\Rightarrow v=4\sqrt{10}\,m/s\\\Rightarrow e=40.40\times {{10}^{-6}}\times 1\times 4\sqrt{10}\\\Rightarrow e =228.53\times {{10}^{-6}}\,V\\\text{Magnitude of difference-}\\\therefore 404\times {{10}^{-6}}V-228.53\times {{10}^{-6}}V=1.7\times {{10}^{-4}}\,V$

A coil having a number of turns $2$0 , self-inductance $L=4\,H$ and current varying with the equation $I=I\sin (\omega t)$ here $\omega $ is constant and $t$ is time. Change in flux in a coil will be- 

The flux can change over time as a result of the current flowing through the coil.

 $e=-L\dfrac{di}{dt}\\\Rightarrow e=-4\dfrac{d(I\sin (\omega t))}{dt}\\\Rightarrow e=-4I\omega \cos (\omega t)\\\Rightarrow e=-N\dfrac{d\phi }{dt}\\\Rightarrow e=-20\dfrac{d\phi }{dt}\\\text{Comparing both emf we have-}\\-20\dfrac{d\phi }{dt}=-4I\omega \cos (\omega t)\\\therefore\dfrac{d\phi }{dt}=\dfrac{I\omega \cos (\omega t)}{5}$

If a rod of length $L$ is rotated along one of its ends with an angular frequency $\omega $ in a constant magnetic field $B$. The emf induced $E$ at a particular moment is used in a below circuit to find the current flowing in a circuit.<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/e2c1d4f2-31d9-43ed-8c68-13d8381677b55500302009722995650.png"/> 

With an increase in the angular velocity of the rod or magnetic field, the current flowing through the circuit will increase.

 $\text{Emf in rotating rod-}\\e=\dfrac{1}{2}B\omega {{l}^{2}}\\\text{Equivalent resistance in above circuit-}\\\text{All three resistors of $3\Omega $ are in parallel so equivalent resistance will be-}\\\dfrac{1}{R}=\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3}=1\\\Rightarrow R=1\\\text{Now it is in series with}\,5\Omega\,\text{resistor so-}\\R=1+5=6\Omega\\\text{Current flowing will be-}\\i=\dfrac{e}{R}\\\Rightarrow i =\dfrac{\dfrac{1}{2}B\omega {{l}^{2}}}{R}\\\therefore i =\dfrac{B\omega {{l}^{2}}}{12}$