Question

A solenoid of length $50 \,cm$, having $100$ turns carries a current of $2.5\, A$. The magnetic field at one end of the solenoid is

• A. $3.14 \times 10^{-4}\, T$
• B. $6.28 \times 10^{-4}\, T$
• C. $1.57 \times 10^{-4}\, T$
• D. $9.42 \times 10^{-4}\, T$

Answer 1

Answer: (A)

Here, $I=2.5 \,A$,
$ l=50\,cm $
$=0.50\, m$
and $n=\frac{100}{0.50}=200\, m^{-1}$
$\therefore B=\frac{\mu_{0}nI}{2}=\frac{4\pi\times10^{-7}\times200\times2.5}{2}$
$=3.14 \times10^{-4}\,T $