The current induction at the center of a colenoid of length 2 cm and diamteter 4cm is `3 xx 10^(-3)Wb//m^(2)`. What is the magnetic flux through the cross section of the solenoid ?

Options

A. `2 xx 10^(-6)Wb`
B. `3.8 xx 10^(-6)Wb`
C. `5 xx 10^(-6)Wb`
D. `8 xx 10^(-6)Wb`

Answer

B

Explanation

Area of cross section`pir^(2)=pi xx (2 xx 10^(-2))^(2)=4pi xx 10^(-4)therefore phi=BA=B.pir^(2)=3 xx 10^(-3)xx 4pi xx 10^(-4)=12 pi xx 10^(-7)=3.8 xx 10^(-6)Wb`.