Question

A long solenoid with \(500\) turns per unit length carries a current of \(1.5\) A. The magnetic induction at one of the ends of the solenoid on its axis in tesla is

• A. \(32\times10^{-4}\) T
• B. \(4\times10^{-4}\) T
• C. \(47\times10^{-4}\) T
• D. \(16\times10^{-4}\) T
• E. \(8\times10^{-4}\) T

Hint:

For a long solenoid: \[ B_{\text{center}}=\mu_0 nI,\qquad B_{\text{end}}=\frac{1}{2}\mu_0 nI \]

Answer 1

The Correct Option is D

Magnetic field at the end of a long solenoid: \[ B=\frac{1}{2}\mu_0 n I \] Given: \[ n=500\ \text{turns/m},\quad I=1.5\text{ A},\quad \mu_0=4\pi\times10^{-7} \] So, \[ B=\frac{1}{2}\cdot 4\pi\times10^{-7}\cdot 500\cdot 1.5 \] \[ B=2\pi\times10^{-7}\cdot 750 \] \[ B=1500\pi\times10^{-7} \approx 4.71\times10^{-4}\text{ T} \] This direct value matches: \[ \boxed{4.7\times10^{-4}\text{ T}} \] which corresponds to option (C), not (D).