The charge which will flow through a galvanom | vedclass

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Question

(A) The total charge $q$ flowing through a circuit due to a change in magnetic flux $\Delta \phi$ is given by the formula $q = \frac{\Delta \phi}{R_{t

Vedclass · Accepted Answer

$63$

Vedclass · Answer

(A) The total charge $q$ flowing through a circuit due to a change in magnetic flux $\Delta \phi$ is given by the formula $q = \frac{\Delta \phi}{R_{total}}$.Here,$R_{total} = R_{galvanometer} + R_{coil} = 200 \ \Omega + 400 \ \Omega = 600 \ \Omega$.The change in flux is $\Delta \phi = N A \Delta B$,where $N = 1000$ and $A = \pi r^2$.The diameter is $20 \ mm$,so the radius $r = 10 \ mm = 0.01 \ m$.Thus,$A = \pi 	imes (0.01)^2 = \pi 	imes 10^{-4} \ m^2$.The change in magnetic field $\Delta B = |0 - 0.012| = 0.012 \ T$.Substituting these values into the charge formula:$q = \frac{1000 	imes \pi 	imes 10^{-4} 	imes 0.012}{600}$$q = \frac{1000 	imes 3.14159 	imes 10^{-4} 	imes 0.012}{600}$$q = \frac{3.14159 	imes 0.012}{600} 	imes 1000 = \frac{0.037699}{600} 	imes 1000 \approx 0.0628 	imes 10^{-3} \ C = 62.8 \ \mu C$.Rounding to the nearest value,we get $q \approx 63 \ \mu C$.

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