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# The magnetic field in the cylinder region shown in the figure increases at a constant rate of 20.0mT/s. each side of the square loop abcd and defa has a length of 1.00cm and a resistance of 4.00$\Omega$. Find the current (magnitude and sense) in the wire and ifA. The switch ${{S}_{1}}$is closed but ${{S}_{2}}$ is open,B.${{S}_{1}}$ is open but ${{S}_{2}}$ is closed,C. both ${{S}_{1}}$and ${{S}_{2}}$ are openD. both ${{S}_{1}}$and ${{S}_{2}}$are closed

Last updated date: 20th Jun 2024
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Hint: In order to find current we have to find the emf and the resistance of the circuit in (a) current will flow in edaf in (b) current will flow in abcd in (c) no current will flow in (d) circuit will act as a balanced Wheatstone bridge hence no current will flow along ad.
Formula used:
$\phi =B.A$
\begin{align} & e=\dfrac{d\phi }{dt} \\ & e=emf \\ \end{align}

A) ${{S}_{1}}$ Closed and ${{S}_{2}}$ is open

Formula for the flux = $\phi =B.A\cos \theta$
Here the θ will be zero because magnetic field is perpendicular
$\phi =B.A$
Where,
$\phi =$ Flux
B = magnetic field
A = area
Now emf,
\begin{align} & \Rightarrow e=\dfrac{d\phi }{dt} \\ & \Rightarrow e=A\dfrac{dB}{dt} \\ & \therefore e={{l}^{2}}\times \dfrac{dB}{dt}....(1) \\ \end{align}
It is given that
\begin{align} & \Rightarrow l=1cm \\ & \Rightarrow l=1\times {{10}^{-2}}m \\ & \Rightarrow \dfrac{db}{dt}=20mT/s \\ & \therefore \dfrac{db}{dt}=20\times {{10}^{-3}}T/s \\ \end{align}
Substituting value in equation (1)
\begin{align} & \Rightarrow e={{\left( 1\times {{10}^{-2}} \right)}^{2}}\times 20\times {{10}^{-3}} \\ & \Rightarrow e={{10}^{-4}}\times 2\times {{10}^{-2}} \\ & \therefore e=2\times {{10}^{-6}} \\ \end{align}
Now we know that current
$I=\dfrac{e}{R}....\left( 2 \right)$
Here it is given that resistance is $4\Omega$ on every side hence
\begin{align} & \Rightarrow R=4+4+4+4 \\ & \therefore R=16\Omega \\ \end{align}
Hence the current
\begin{align} & \Rightarrow I=\dfrac{2\times {{10}^{-6}}}{16} \\ & \therefore I=1.25\times {{10}^{-7}}A \\ \end{align}
B.${{S}_{1}}$ is open but ${{S}_{2}}$ is closed,

Current will form a circuit abcd since all values are the same; the magnitude of the current will also be the same. Hence
$I=1.25\times {{10}^{-7}}A$
C. both ${{S}_{1}}$and ${{S}_{2}}$ are open
If both ${{S}_{1}}$and ${{S}_{2}}$ are open then the circuit will not complete hence current along and is zero.
D. both ${{S}_{1}}$and ${{S}_{2}}$are closed

Since all sides have equal resistance, the circuit can act as a balanced Wheatstone bridge.
Formula for checking balanced Wheatstone bridge.
\begin{align} & \Rightarrow \dfrac{{{R}_{1}}}{{{R}_{2}}}=\dfrac{{{R}_{3}}}{{{R}_{4}}}=1 \\ & \therefore \dfrac{4}{4}=\dfrac{4}{4}=1 \\ \end{align}
Hence the circuit will act as a balanced Wheatstone bridge hence current flow along ad is zero.

Note:
We have to remember two points first it is given that resistance is all four sides so don’t forget to consider resistance and in (d) we have to check for Wheatstone bridge otherwise it will lead us to the wrong answer.