Question

A solenoid of length $0.6\, m$ has a radius of $2\, cm$ and is made up of $600$ turns If it carries a current of $4 \,A$ , then the magnitude of the magnetic field inside the solenoid is

• A. $6.024 \times 10^{-3} \, T$
• B. $8.024 \times 10^{-3}\, T$
• C. $5.024 \times 10^{-3} \, T$
• D. $7.024 \times 10^{-3}\, T$

Answer 1

Answer: (C)

Here, $n=\frac{600}{0.6}=1000$ turns / m, $ I=4A$
$l=0.6\,m$, $r=0.02\,m$
$\because \frac{l}{r}=30$ i.e. $l >>r $
Hence, we can use long solenoid formula, then
$\therefore B=\mu_{0}nI=4\pi\times10^{-7}\times10^{3}\times4=50.24\times10^{-4}$
$=5.024\times10^{-3}\, T $