A long solenoid of 50 cm length having 100 turns carries a current of 2.5 A. The magnetic field at the centre

Question

A long solenoid of 50 cm length having 100 turns carries a current of 2.5 A. The magnetic field at the centre of the solenoid is: $(\mu_0 = 4\pi 	imes 10^{-7} 	ext{ T m A}^{-1})$

Accepted Answer

1. **Formula:** The magnetic field $B$ inside a long solenoid is given by the formula $B = \mu_0 n I$, where $\mu_0$ is the permeability of free space, $n$ is the number of turns per unit length, and $I$ is the current .
2. **Calculate $n$:** The number of turns per unit length $n$ is calculated as $n = \frac{N}{L}$.
   Given $N = 100$ turns and $L = 50 	ext{ cm} = 0.5 	ext{ m}$.
   $$n = \frac{100}{0.5} = 200 	ext{ turns/m}$$
3. **Substitute Values:**
   $$B = (4\pi 	imes 10^{-7} 	ext{ T m A}^{-1}) 	imes (200 	ext{ m}^{-1}) 	imes (2.5 	ext{ A})$$
   $$B = 4 	imes 3.14 	imes 10^{-7} 	imes 500$$
   $$B = 6280 	imes 10^{-7} 	ext{ T}$$
   $$B = 6.28 	imes 10^{-4} 	ext{ T}$$

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