A metallic square wire loop of side 10, cm an | vedclass

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(C) The resistor network connected across $P$ and $O$ forms a balanced Wheatstone bridge. The four resistors of $3\,\Omega$ each form the arms of the

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$2 	imes 10^{-2}\, m/s$,clockwise direction

Vedclass · Answer

(C) The resistor network connected across $P$ and $O$ forms a balanced Wheatstone bridge. The four resistors of $3\,\Omega$ each form the arms of the bridge,and the central $3\,\Omega$ resistor is in the middle. Since the bridge is balanced,no current flows through the central resistor.The equivalent resistance of the network is the parallel combination of two branches,each having two $3\,\Omega$ resistors in series. Thus,$R_{eq} = \frac{(3+3) 	imes (3+3)}{(3+3) + (3+3)} = \frac{6 	imes 6}{12} = 3\,\Omega$.The induced electromotive force (emf) in the moving loop is given by $\varepsilon = B l v_0$.Using Ohm's law for the circuit,$I = \frac{\varepsilon}{R_{eq} + R_{loop}}$,where $R_{loop} = 1\,\Omega$ is the resistance of the square loop.Substituting the values: $10^{-3}\, A = \frac{2\, T 	imes 0.1\, m 	imes v_0}{3\,\Omega + 1\,\Omega}$.$10^{-3} = \frac{0.2 	imes v_0}{4} \Rightarrow 10^{-3} = 0.05 	imes v_0$.$v_0 = \frac{10^{-3}}{0.05} = 2 	imes 10^{-2}\, m/s$.According to Lenz's law,as the loop moves out of the magnetic field,the magnetic flux linked with the loop decreases. To oppose this decrease,the induced current will flow in a direction to create magnetic flux into the page,which is the clockwise direction.

List-$I$	List-$II$
$A$. Magnetic induction	$I$. $MLT^{-2}A^{-2}$
$B$. Magnetic flux	$II$. $ML^2T^{-2}A^{-2}$
$C$. Magnetic permeability	$III$. $ML^0T^{-2}A^{-1}$
$D$. Self inductance	$IV$. $ML^2T^{-2}A^{-1}$

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Match the List-$I$ with List-$II$:
List-$I$ List-$II$
$A$. Magnetic induction $I$. $MLT^{-2}A^{-2}$
$B$. Magnetic flux $II$. $ML^2T^{-2}A^{-2}$
$C$. Magnetic permeability $III$. $ML^0T^{-2}A^{-1}$
$D$. Self inductance $IV$. $ML^2T^{-2}A^{-1}$

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