# A 0.1 m long solenoid has a radius of 0.05 m and 1.5 x 10^4 turns. The current in the solenoid...

## Question:

A 0.1 m long solenoid has a radius of 0.05 m and {eq}1.5 \times 10^4 {/eq} turns. The current in the solenoid changes at a rate of 6.0 A/s. A conducting loop of radius 0.0200 m is placed at the center of the solenoid with its axis the same as that of the original solenoid.

a. What is the magnetic flux through the small loop when the current through the solenoid is 2.5 A?

b. Determine the self-induced emf in the solenoid due to the changing current.

## Induced electromotive force:

When a coil is placed inside a changing magnetic field it induces an electromotive force inside it. The direction of this emf is opposite the direction of the flux because it opposes the cause which is producing it.

From Lenz's law, the induced emf is formulated as:

\begin{align} \color{red}{\varepsilon=-N\frac{\mathrm{d} \phi}{\mathrm{d} t}} \end{align}

Here:

• {eq}\phi {/eq} is the flux.
• {eq}N {/eq} is the number of turns.

Given:

• The length of the solenoid {eq}\begin{align} L=0.1 \ \text{m} \end{align} {/eq}
• The radius of the solenoid {eq}\begin{align} r=0.05 \ \text{m} \end{align} {/eq}
• The number of turns in the solenoid {eq}\begin{align} N=1.5\times10^4 \ \text{turns} \end{align} {/eq}
• The rate of the change in current {eq}\begin{align} \frac{\mathrm{d} I}{\mathrm{d} t}=6.0 \ \text{A/s} \end{align} {/eq}
• The radius of the loop {eq}\begin{align} R=0.02 \ \text{m} \end{align} {/eq}

(a)

The flux produced by the solenoid is going through the loop. Thus the magnetic flux flowing through the loop when the current {eq}I=2.5 \ \text{A} {/eq}, is:

{eq}\begin{align} \phi&=(B)(A)\\ &=(\mu_0nI)(\pi R^2)\\ &=\mu_0\left(\frac{N}{L}\right)I\times\pi R^2\\ &=4\pi\times10^{-7}\ NA^{-2}\times\left(\frac{1.5\times10^4}{0.1\ m}\right)\times2.5\ A \times\pi(0.02\ m)^2\\ &=\color{blue}{5.919\times10^{-4} \ \text{Wb}} \end{align} {/eq}

(b)

Recall that from Lenz's law, the induced emf is formulated as:

{eq}\begin{align} \left|\varepsilon\right|&=\left|N\frac{\mathrm{d} \phi}{\mathrm{d} t}\right|\\ \\ &=N\frac{\mathrm{d} (BA)}{\mathrm{d} t}\\ \\ &=NA\frac{\mathrm{d} B}{\mathrm{d} t}\\ \\ &=NA\frac{\mathrm{d} }{\mathrm{d} t}(\mu_0 nI)\\ \\ &=\mu_0\left(\frac{N^2}{L}\right)(\pi r^2)\frac{\mathrm{d} I}{\mathrm{d} t}\\ \\ &=4\pi\times10^{-7}\ NA^{-2} \times\frac{(1.5\times10^4)^2}{0.1\ m}\times\pi(0.05\ m)^2\times6.0\ A/s\\ \\ &=\color{blue}{133.23 \ \text{V}} \end{align} {/eq}