Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

A solenoid is 2.7-m long, has a radius of 0.85 cm, and has 600 turns. It carries a current I of 2.5 A. What is the magnitude of the magnetic field B inside the solenoid and far from either end?

$\begin{array}{1 1} 0.007\;mT \\ 7\;mT \\ 0.07\;mT \\ 0.7 \;mT \end{array}$

Can you answer this question?

1 Answer

0 votes
The formula for the field inside the solenoid is $B = \mu_0 n I$, where $n = \large\frac{N}{L}$ or the turn density and $I$ is the current.
Given that number of turns, $N = 600$ and the length of the solenoid, $L = 2.7m \rightarrow n = \large\frac{600}{2.7} $$ \approx 222.22 \;m^{-1}$
$\Rightarrow B = 4\pi \times 10^{-7} N/A^{-2} \times 222.22 m^{-1} \times 2.5A \approx 0.7mT$
answered Mar 14, 2014 by balaji.thirumalai

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App