Question

A solenoid has $100$ turns/meter and is carrying a current of $.4A$. The radius of coil is $8cm$. The magnitude of magnetic induction at a point $50cm$ on the axis from centre of solenoid is

• A. $20.1\times {10}^{-6}T$
• B. $12\times {10}^{-6}T$
• C. $3.1\times {10}^{-4}T$
• D. $5\times {10}^{-6}T$

Answer 1

Answer: (C)

$3.1\times {10}^{-4}T$

given : magnitude of magnetic feild inside the solenoid remains constant

solution : we know the formula for magnetic feild inside the solenoid , given by

$B=\frac{\mu oIN}{L}$

WHERE ,

I is the current flowing in solenoid

N is the number of turns per m

L is the length of solenoid i.e. diameter of solenoid in meter

putting I= 0.4

$\mu o=4\pi x{10}^{-}7$ N = 100, L =0.16 ,

we get

B ={$4\pi x{10}^{-}7$ X 0.4 X 100}/{0.16}

$B=3.1$ x ${10}^{-}4T$

Hence the correct OPT : C