Magnetic Field at the End of Solenoid – Rankers Physics

Force Acting on Current Carrying Conductor: Practice Problem & Solution

A long solenoid having number of turns per unit length 200 carries a current of \(2.5 \text{ A}\), the magnetic field at the end of the solenoid is
\(6.28 \times 10^{-4} \text{ T}\)
\(3.14 \times 10^{-4} \text{ T}\)
\(6.28 \times 10^{-5} \text{ T}\)
\(3.14 \times 10^{-5} \text{ T}\)

Solution Explained:

To solve this problem, we apply the core principles of Force Acting on Current Carrying Conductor. Understanding the underlying formula is key to arriving at the correct answer below:

The magnetic field at the end of a long solenoid is \(B_{\text{end}} = \frac{1}{2} \mu_0 n I\). Substituting the given values: \(B_{\text{end}} = \frac{1}{2} (4\pi \times 10^{-7}) (200) (2.5) = 3.14 \times 10^{-4} \text{ T}\).

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