Question 26 of 276
The magnetic field inside a 200 turns solenoid of radius 10 cm is . If the solenoid carries a current of 0.29 A , then the length of the solenoid is ________ .
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Answer: 8
Assuming long solenoid
$\begin{aligned}
& \mathrm{B}=\mu_0\left(\frac{\mathrm{~N}}{\ell}\right) \mathrm{i} \\
& \ell=\frac{\mu_0 \mathrm{Ni}}{\mathrm{~B}}=\frac{\left(4 \pi \times 10^{-7}\right)(200)(0.29)}{2.9 \times 10^{-4}} \mathrm{~m} \\
& =8 \pi \mathrm{~cm}
\end{aligned}$
$\begin{aligned}
& \mathrm{B}=\mu_0\left(\frac{\mathrm{~N}}{\ell}\right) \mathrm{i} \\
& \ell=\frac{\mu_0 \mathrm{Ni}}{\mathrm{~B}}=\frac{\left(4 \pi \times 10^{-7}\right)(200)(0.29)}{2.9 \times 10^{-4}} \mathrm{~m} \\
& =8 \pi \mathrm{~cm}
\end{aligned}$
