Question

A solenoid is 0.30 m long with 103 turns per meter and carries a current of 5.0 A. what is

the magnitude of the magnetic field at the center of this solenoid?

Answer 1

we know that magnetic field at center of solenoid is given by - $B= {\mu_0 NI}$

N = $\dfrac {n}{l}$ now putting the value of N in above formula , we get

$B=\dfrac {\mu_0 nI}{L}$

${\mu_0}=4{\pi}\times 10^{-7}$ H/m

$n=$ 103

$L=$ 0.30 m

$I=5.0 A$

now putting all the above equation ,

$B=\dfrac{4{\pi}\times 10^{-7}\times 103 \times 5.0}{0.30}$

$B=2.156\times 10^{-3}$ T