Ampere's Circuital Law: Practice Problem & Solution
A long solenoid of $50text{ cm}$ length having $100$ turns carries a current of $2.5text{ A}$. The magnetic field at the centre of the solenoid is: ($\mu_0 = 4\pi \times 10^{-7}\text{ T m A}^{-1}$) (2020)
Solution Explained:
To solve this problem, we apply the core principles of Ampere's Circuital Law. Understanding the underlying formula is key to arriving at the correct answer below:
The number of turns per unit length is $n = \frac{N}{L} = \frac{100}{0.5\text{ m}} = 200\text{ turns/m}$. Using the formula $B = \mu_0 n I$, we substitute $\mu_0 = 4\pi \times 10^{-7}$, $n = 200$, and $I = 2.5\text{ A}$ to get $B = 6.28 \times 10^{-4}\text{ T}$.
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