Question

64. II A solenoid inductor has an emf of $0.20\text{V}$when the current through it changes at the rate $10.0\text{A}/\text{s}$ . A steady current of 0.10 A produces a flux of $5.0\mu \text{Wb}$per turn. How many turns does the inductor have?

Short answer

The number of turns in the solenoid are 39

Step-by-step solution

Step 1. Given information

The Magnetic flux,$B=5\mu \text{Wb}/\text{turn}=5\times {10}^{-6}\text{Wb}/\text{turn}$

The current,$I=0.10\text{A}\mu =4\pi \times {10}^{-7}$

Step 2. Explanation

Flux in a solenoid is given by

$B=\mu nI$

Here is magnetic flux, is current and is number of turns.

From equation (1),

$\begin{array}{l}n=\frac{B}{\mu I}\\ n=\frac{5\times {10}^{-6}}{4\times 3.14\times {10}^{-7}\times 0.10}\\ n=39.8\\ n\approx 39\end{array}$

The number of turns in the solenoid are 39