Question

If the field strength in the solenoid for the bell in part (a) is 2.00x10 -5 T and there are 6

loops (exact), then find the current sent out by the battery. The length is 12.0cm. [4 marks]

Answer 1

$We \ know \ that \ magnetic\ field\ inside\ solenoid\ is\ given\ by\ equation \ , B= \dfrac{{{\mu_O}N}I}{L}$

Magnetic field inside a solenoid $B=2.00\times10^{-5}T$

$Number\ of\ turns\ N=6 \ loops$

$Length\ L=12.0\ cm$ =0.12 m

$Absolute\ premitivitty\ {\mu}_o$ $=4{\pi}\times10^{-7}$ H/m

$2.00\times10^{-5}=\dfrac{4{\pi}\times10^{-7}\times 6\times I}{0.12}$

$I=\dfrac{2.00\times10^{-5}\times 0.12}{4{\pi}\times10^{-7}\times 6}=0.318 \ A$