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 x ${10}^3$ T
• B. 8.024 x ${10}^3$ T
• C. 5.024 x ${10}^3$ T
• D. 7.024 x ${10}^3$ T

Answer 1

Answer: (C)

5.024 x ${10}^3$ T

Here,

$I=4Al=0.6m,r=0.02m$ $\because$ $\frac{l}{r}$ = 30 i.e. l >> r

$n=\frac{600}{0.6}=1000turns/m,$

Hence, we can use long solenoid formula, then

$\therefore$ $B={\mu }_{0}nI=4\pi$ x ${10}^{-7}$ x ${10}^3$ x 4 = 50.24 x ${10}^{-4}$ = $5.024$ x ${10}^3$ T