Question

A 100-turn, the 2.0-cm-diameter coil is at rest with its axis vertical. A uniform magnetic field $60°$away from vertical increases from $0.50\mathrm{T}$to $1.50\mathrm{T}$in $0.60\mathrm{s}$. What is the induced emf in the coil?

Short answer

$\epsilon =26\mathrm{mV}$is the current in the loop.

Step-by-step solution

Introduction

When a conductor is put in a fluctuating magnetic field, "an electrode material effect is generated," according to Faraday's original electromagnetic hypothesis. A current is induced when ever a conductor circuit is completed, and this is called as an induced current.

Simplification

The electromotive force is the charge of magentic flux inside of the loop, as determined by Faraday's law, and it is defined by equation $(30.14)$in the form

Equation $1$

$\epsilon =\left|\frac{d{\Phi }_{\mathrm{m}}}{dt}\right|$

With ${\mathrm{\Phi }}_{\mathrm{m}}$is the flux and through loop, which is the amount of magnetic field that flows through a loop of area $A$, is the flux through the loop, which is given by

${\Phi }_{\mathrm{m}}=NBA\cos \theta$

Let's plug this ${\mathrm{\Phi }}_{\mathrm{m}}$expression into equation ($1$) to get $\epsilon$.

Equation $2$

$\epsilon =N\frac{d\left(BA\cos \theta \right)}{dt}=\left(NA\cos \theta \right)\frac{dB}{dt}$

Calculate the area of loop

The loop's area is computed by

$A=\pi {\left(\frac{d}{2}\right)}^2=\pi {\left(\frac{0.02\mathrm{m}}{2}\right)}^2=3.14\times {10}^{-4}{\mathrm{m}}^2$

In the time $dt=0.60$, the strong magnetic changes from $0.50$to $1.5$, then we get this changes over the years by

$\frac{dB}{dt}=\frac{1.5\mathrm{T}-0.50\mathrm{T}}{0.60\mathrm{s}}=1.67\mathrm{T}/\mathrm{s}$

To be get $\epsilon$, we plug the values of $N$, $A$, $\theta \text{and}(dB/dt)$with Equation (2).

$\begin{array}{r}\epsilon =NA\cos \theta \frac{dB}{dt}\\ =\left(100\right)\left(3.14\times {10}^{-4}{\mathrm{m}}^2\right)\cos {60}^{\circ }(1.67\mathrm{T}/\mathrm{s})\\ =26\times {10}^{-3}\mathrm{V}\\ =26\mathrm{mV}\end{array}$