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

Magnetic flux Quiz 2

physics

 A square-shaped coil is perpendicularly placed in a uniform magnetic field with a value of $1$ Tesla. The coil's dimensions change at a rate of $\text{1cm}/\text{sec}$. The electromotive force $\left( EMF \right)$ induces in the coil whose side of the square is $5$ metre. 

Calculate change in flux to predict the magnitude of emf in the coil. Find derivative of flux to the value of emf. 

  $\phi =B.A \\ \Rightarrow\phi =B\times {{\ell }^{2}} \\ \Rightarrow emf=\left| \dfrac{d\phi }{dt} \right|\\ \Rightarrow emf ={B_2}\ell \dfrac{d\ell }{dt}\\ \Rightarrow emf =1\times 5\times 0.01\\ \therefore emf =0.05\,V$

Which of the following graphs suit’s the given condition the most? A horizontal bar of length $\ell $ is executing a simple harmonic motion $\vec{y}={{y}_{o}}\sin \omega t\hat{k}$. If there is a magnetic field $B={{B}_{o}}\hat{k}$, then the rate of change of $emf$ $E$ is 

$\text{Velocity }V=\dfrac{dy}{dt}$, where $y$ is the displacement of performing $SHM,y={{y}_{0}}\sin \omega t$. 

  &nbsp;$y={{y}_{o}}\sin \omega t \\ \text{Differentiating with respect to time gives velocity} \\V=\dfrac{dy}{dt}=\omega {{y}_{o}}\cos \omega t \\ \text{As }\!\!~\!\!\text{ }emt\text{ induced is} \\E=B\ell V \\ \Rightarrow E=B\ell \omega {{y}_{o}}\cos \omega t \\ \text{Rate of change of }emt\text{ is} \\ \dfrac{dE}{dt}=-B\ell {{\omega }^{2}}{{Y}_{o}}Sin\omega t \\ \text{Thus, maximum value of }\!\!~\!\!\text{ }E\text{ is }B\ell {{\omega }^{2}}{{y}_{o}}\text{.}$

 A square with side $a$ is put in a magnetic field $B$ and extended outwards at a rate of  $2\text{ m/s}$ along one of the sides. A similar square is subjected to similar conditions, but it is reduced inwards at a rate of $2\text{ m/s}$ along one of the edges. If magnetic field is $B$ perpendicular to these squares, the $Emf$ induced in case (1) and case (2) are

$V=\dfrac{d\phi }{dt}=\dfrac{d}{dt}\left( \vec{B}.\vec{A} \right)$ where $B$ is the magnetic field $A$ is the area of cross-section. Here the induced Emf does not depend on the magnetic field or area, instead it just depends on the rate of change of the said quantities. 

 $\text{The Emf is induced according to the rate of change of area}\text{.} \\ \text{Rate of change of area in both cases is }\\ \dfrac{dA}{dt}=a.\dfrac{da}{dt},a\text{ is side of square} \\ \Rightarrow a.2=2a \\ \text{Thus Emf induced in both cases is} \\V=\dfrac{d\phi }{dt}=\dfrac{d}{dt}\left( \vec{B}.\vec{A} \right) \\ \Rightarrow \dfrac{BdA}{dt} \\ \Rightarrow B.\left( 2a \right) \\ \therefore E=2aB \\ \text{Thus }{{E}_{1}}={{E}_{2}}=E=2aB$

A rectangular conductive coil of dimension  $\left( \ell \times b \right)$ moves at a constant velocity $v$.It enters from region $1$ to $2$, where there is a uniform magnetic field inward to the plane exits, and then it enters region $3$, where there is no magnetic field. The magnitude of the induced emf during its motion in region $2$ is $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/2b4d8fc7-b71f-43f6-8fc4-b79df6abdd845665509184743919159.png"/>

Magnitude of emf is a conducting rod travelling with value its $V$ in perpendicular magnetic field is $BLV$ and $L$ is length perpendicular to velocity. 

 $\text{In region }\left( 2 \right) \\ \text{induced Emf}=BLV \\ \text{as the length perpendicular to velocity }V \\ \text{Magnetic field is B,}\\ \text{ so Emf will be induced across it}\text{.} \\ \text{Thus, induced Emf}=BbV$

A conducting coil having resistance $2\Omega $ experience a flow of current of $3$ ampere due to continuous change in magnetic field. Determine the change in flux for time interval $2-4$ seconds.<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/7f825ef7-4074-4008-a5d7-774e87d29d631849296866047840268.png"/>

 In closed circuit $V=IR$ and $v=\dfrac{d\theta }{dt},$analyze using these formulae. 

 $\text{The induced Emf }'V'\text{ due to changing magnetic field }'B'\text{ is} \\V=\dfrac{\Delta \phi }{\Delta t} \\ \text{This voltage, by Ohm }\!\!'\!\!\text{ s law} \\V=IR \\ \Rightarrow  IR=\dfrac{\Delta \phi }{\Delta t} \\ \Rightarrow 3\times 2=\dfrac{\Delta \phi }{4-2} \\ \therefore \Delta \phi =12\text{ Weber}$