Magnetic Field at the End of Solenoid – Rankers Physics
Topic: Magnetic Effects of Current
Subtopic: Force Acting on Current Carrying Conductor

Magnetic Field at the End of Solenoid

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:

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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