A \( 0.5\,\mathrm{m} \) long solenoid has \( 500 \) turns and has a flux density of \( 3.14\times10^{-3}\,\mathrm{Wb/m^{2}} \) at its centre. The current in the solenoid is:
[Given: \( \mu_{0} = 4\pi\times10^{-7}\,\mathrm{Tm/A} \)]
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 2.5\,\mathrm{A} \)
The field inside a long solenoid is \( B = \mu_0 nI \), where \( n \) is the number of turns per unit length.
\( n = N/l = 500/0.5\,\mathrm{m} = 1000\,\mathrm{m^{-1}} \)
\( B = 3.14\times10^{-3}\,\mathrm{Wb/m^{2}} = 3.14\times10^{-3}\,\mathrm{T} \)
\( \displaystyle I = \frac{B}{\mu_0 n} = \frac{3.14\times10^{-3}\,\mathrm{T}}{4\pi\times10^{-7}\,\mathrm{Tm/A}\times1000\,\mathrm{m^{-1}}} \)
\( \displaystyle I = \frac{3.14\times10^{-3}}{1.256\times10^{-3}}\,\mathrm{A} = 2.5\,\mathrm{A} \)