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(B) The magnetic field $B$ inside a long solenoid is given by $B = \mu_0 n I$, where $n$ is the number of turns per unit length and $I$ is the current

Vedclass · Accepted Answer

$0.4$

Vedclass · Answer

(B) The magnetic field $B$ inside a long solenoid is given by $B = \mu_0 n I$, where $n$ is the number of turns per unit length and $I$ is the current.Given: $n = 100 	ext{ turns/cm} = 10^4 	ext{ turns/m}$, $I = \frac{4}{\pi} \,A$, and $\mu_0 = 4\pi 	imes 10^{-7} \,T \cdot m/A$.$B = (4\pi 	imes 10^{-7}) 	imes (10^4) 	imes (\frac{4}{\pi}) = 16 	imes 10^{-3} \,T$.The magnetic flux $\phi_B$ linked with the coil of $N$ turns and area $A$ is $\phi_B = N B A$.Given: $N = 200$, $A = 25 \,cm^2 = 25 	imes 10^{-4} \,m^2$.When the current is reversed, the magnetic field changes from $B$ to $-B$. The change in flux is $\Delta \phi_B = N(B - (-B))A = 2NBA$.The induced emf $e$ is given by Faraday's law: $e = \left| \frac{\Delta \phi_B}{\Delta t} ight| = \frac{2NBA}{\Delta t}$.Substituting the values: $e = \frac{2 	imes 200 	imes (16 	imes 10^{-3}) 	imes (25 	imes 10^{-4})}{0.04} = \frac{400 	imes 16 	imes 10^{-3} 	imes 25 	imes 10^{-4}}{0.04} = \frac{160000 	imes 10^{-7}}{0.04} = \frac{0.016}{0.04} = 0.4 \,V$.

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