Q.26 A solenoid of length \(\pi\) m and 5 cm in diameter has a winding of 1000 turns and carries a current of 5A. Calculate the magnetic field at its centre along the axis.
Best answer
Answer: \(B = 2 \times 10^{-3}\) T
Since the length (\(\pi\) m) is much greater than the diameter (5 cm), the solenoid can be treated as a long solenoid. Number of turns per unit length
\[n = \frac{N}{l} = \frac{1000}{\pi} \text{ turns/m}\]Magnetic field at the centre along the axis:
\[B = \mu_0 n I = 4\pi \times 10^{-7} \times \frac{1000}{\pi} \times 5 = 2 \times 10^{-3} \text{ T}\]The field is directed along the axis of the solenoid.
Log in or create a free account to answer this question.
