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 x ${10}^4$T
• B. 6.28 x ${10}^4$T
• C. 1.57 x ${10}^4$ T
• D. 9.42 x ${10}^4$ T

Answer 1

Answer: (A)

3.14 x ${10}^4$T

Here,

$I=2.5A,l=50cm=0.50mandn=\frac{100}{0.50}=200m^{-1}$

$\therefore$ $B=\frac{{\mu }_{0}nI}{2}=\frac{4\pi \times {10}^{-7}\times 200\times 2.5}{2}$ = 3.14 x ${10}^4$T