A flat coil of 500 turns,each of area 50 ,cm^ | vedclass

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(C) Given:Number of turns $N = 500$Area $A = 50 \,cm^2 = 50 	imes 10^{-4} \,m^2 = 5 	imes 10^{-3} \,m^2$Magnetic field $B = 0.14 \,Wb/m^2$Angular sp

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(C) Given:Number of turns $N = 500$Area $A = 50 \,cm^2 = 50 	imes 10^{-4} \,m^2 = 5 	imes 10^{-3} \,m^2$Magnetic field $B = 0.14 \,Wb/m^2$Angular speed $\omega = 150 \,rad/s$Internal resistance $r = 5 \,\Omega$External resistance $R = 10 \,\Omega$The peak induced $e.m.f.$ is given by $\varepsilon_{\max} = N B A \omega$.Substituting the values:$\varepsilon_{\max} = 500 	imes 0.14 	imes (5 	imes 10^{-3}) 	imes 150$$\varepsilon_{\max} = 500 	imes 0.14 	imes 0.005 	imes 150 = 52.5 \,V$.The total resistance of the circuit is $R_{total} = R + r = 10 \,\Omega + 5 \,\Omega = 15 \,\Omega$.The peak current $I_{\max}$ is given by $I_{\max} = \frac{\varepsilon_{\max}}{R_{total}}$.$I_{\max} = \frac{52.5}{15} = 3.5 \,A$.

Column $I$	Column $II$
$(A)$ $A$ charged capacitor is connected to the ends of the wire	$(p)$ $A$ constant current flows through the wire
$(B)$ The wire is moved perpendicular to its length with a constant velocity in a uniform magnetic field perpendicular to the plane of motion	$(q)$ Thermal energy is generated in the wire
$(C)$ The wire is placed in a constant electric field that has a direction along the length of the wire	$(r)$ $A$ constant potential difference develops between the ends of the wire
$(D)$ $A$ battery of constant emf is connected to the ends of the wire	$(s)$ Charges of constant magnitude appear at the ends of the wire

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