A coil of 40 Omega resistance has 100 turns | vedclass

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(D) The total resistance of the circuit is $R_{total} = R_{coil} + R_{ammeter} = 40 \ \Omega + 160 \ \Omega = 200 \ \Omega$. The charge $q$ induced in

Vedclass · Accepted Answer

$0.566$

Vedclass · Answer

(D) The total resistance of the circuit is $R_{total} = R_{coil} + R_{ammeter} = 40 \ \Omega + 160 \ \Omega = 200 \ \Omega$. The charge $q$ induced in the coil is given by $q = \frac{\Delta \phi}{R_{total}} = \frac{N A B}{R_{total}}$,where $A = \pi r^2$. Given: $N = 100$,$r = 6 \ mm = 6 	imes 10^{-3} \ m$,$q = 32 \ \mu C = 32 	imes 10^{-6} \ C$. Substituting the values: $32 	imes 10^{-6} = \frac{100 	imes \pi 	imes (6 	imes 10^{-3})^2 	imes B}{200}$. $32 	imes 10^{-6} = \frac{100 	imes 3.1416 	imes 36 	imes 10^{-6} 	imes B}{200}$. $32 = \frac{11309.76 	imes B}{200}$. $32 = 56.5488 	imes B$. $B = \frac{32}{56.5488} \approx 0.566 \ T$.

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