Question

A solenoid with 100 turns in .01m, and a cross-sectional area of $20c{m}^2$ has a current going from zero to 4 Amps in 1 second.
What is the magnitude of the induced emf?

• A. $.032\pi V$
• B. $.0032\pi V$
• C. $3.2\pi V$
• D. $32\pi V$

Answer 1

The correct option is B $.0032\pi V$

Magnitude of induced emf is given by ,

$e=Nd\varphi /dt=d\left(NBA\right)/dt$ ,

where $N=$ no. of turns ,

$A=$ cross-sectional area of solenoid ,

$B=$ magnetic field ,

given $N=100,A=20c{m}^2=20\times {10}^{-4}{m}^2,I_0=0A,I_1=4A,t=1s,l=0.01m$ ,

initial magnetic flux ${\varphi }_0=N{B}_{0}A=N{\mu }_0\left(N/l\right)I_{0}A=0Wb$ ,

final magnetic flux ${\varphi }_1=N{B}_{1}A=N{\mu }_0\left(N/l\right)I_{1}A=100\times 4\pi \times {10}^{-7}\times \left(100/0.01\right)\times 4\times 20\times {10}^{-4}=320\pi \times {10}^{-5}Wb$ ,

$d\left(NBA\right)=320\pi \times {10}^{-5}Wb$ ,

therefore $e=320\pi \times {10}^{-5}/1=0.0032\pi V$