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Question

(B) The number of turns per unit length $n = 15 \; 	ext{turns/cm} = 1500 \; 	ext{turns/m}$.The area of the small loop is $A = 2.0 \; 	ext{cm}^2 = 2

Vedclass · Accepted Answer

$7.54 	imes 10^{-6} \; V$

Vedclass · Answer

(B) The number of turns per unit length $n = 15 \; 	ext{turns/cm} = 1500 \; 	ext{turns/m}$.The area of the small loop is $A = 2.0 \; 	ext{cm}^2 = 2.0 	imes 10^{-4} \; 	ext{m}^2$.The change in current $\Delta I = 4.0 \; 	ext{A} - 2.0 \; 	ext{A} = 2.0 \; 	ext{A}$.The time interval $\Delta t = 0.1 \; 	ext{s}$.The magnetic field inside a long solenoid is $B = \mu_0 n I$.The magnetic flux through the loop is $\phi = BA = \mu_0 n I A$.According to Faraday's law, the induced $emf$ is $e = \left| \frac{d\phi}{dt} ight| = \mu_0 n A \left( \frac{dI}{dt} ight)$.Substituting the values:$e = (4\pi 	imes 10^{-7} \; 	ext{T m/A}) 	imes (1500 \; 	ext{m}^{-1}) 	imes (2.0 	imes 10^{-4} \; 	ext{m}^2) 	imes \left( \frac{2.0 \; 	ext{A}}{0.1 \; 	ext{s}} ight)$.$e = (4 	imes 3.14159 	imes 10^{-7}) 	imes 1500 	imes 2.0 	imes 10^{-4} 	imes 20$.$e = 7.5398 	imes 10^{-6} \; 	ext{V} \approx 7.54 	imes 10^{-6} \; 	ext{V}$.

$A$ long solenoid with $15$ turns per $cm$ has a small loop of area $2.0 \; cm^{2}$ placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from $2.0 \; A$ to $4.0 \; A$ in $0.1 \; s$, what is the induced $emf$ in the loop while the current is changing?

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