Question

A long solenoid has 200 turns per \(\mathrm{cm}\) and carries a current of $2.5 \mathrm{Amp}$. The mag. field at its centre is tesla. (a) \(\pi \times 10^{-2}\) (b) \(2 \pi \times 10^{-2}\) (c) \(3 \pi \times 10^{-2}\) (d) \(4 \pi \times 10^{-2}\)

Short answer

The magnetic field at the center of the solenoid is (a) \(\pi \times 10^{-2} \mathrm{T}\).

Step-by-step solution

Convert turns per centimeter to turns per meter

\ First, we will need to convert the given number of turns per centimeter to turns per meter. Recall that there are 100 centimeters in 1 meter. So, \(n = 200 \text{ turns/cm} \times 100 \text{ cm/m} = 20000 \text{ turns/m}\).

Recall the formula for the magnetic field inside a long solenoid

\ The formula for the magnetic field inside a long solenoid is \(B = \mu_0 n I\), where \(B\) is the magnetic field, \(\mu_0\) is the vacuum permeability constant (\(4\pi \times 10^{-7} \text{ T m/A}\)), \(n\) is the number of turns per meter, and \(I\) is the current.

Plug in the values and solve for the magnetic field

\ Now, plug the given values into the formula and solve for the magnetic field: \(B = (4\pi \times 10^{-7} \text{ T m/A}) \times (20000 \text{ turns/m}) \times (2.5 \mathrm{A})\).

Calculate the magnetic field

\ Using a calculator, compute the magnetic field: \(B = (4\pi \times 10^{-7}) \times (20000) \times (2.5)\) \(B = \pi \times 10^{-2} \mathrm{T}\).

Compare the result to the given options

\ Our calculated magnetic field is \(\pi \times 10^{-2} \text{ T}\), which matches option (a). Therefore, the correct answer is (a) \(\pi \times 10^{-2} \text{ T}\).