Question

A spherical balloon with a volume of $2.5\mathrm{L}$is in a $45\mathrm{mT}$uniform, vertical magnetic field. A horizontal elastic but conducting wire with $2.5\Omega$resistance circles the balloon at its equator. Suddenly the balloon starts expanding at $0.75\mathrm{L}/\mathrm{s}$. What is the current in the wire $2.0\mathrm{s}$later?

Short answer

The current in wire is$I=7.5\times {10}^{-5}\mathrm{A}.$

Step-by-step solution

Step: 1 Magnetic Field:

The magnetic field is uniform in this situation, even though the loop area varies as the coil tightens. The induced emf, as defined by Faraday's law, is the change in magnetic flux inside the loop, and it is provided by equation.

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

Where ${\Phi }_{\mathrm{m}}$does the flux through the loop, which is the amount of magnetic field that travels across an area looplocalid="1648967328330" $A$, originate from

${\Phi }_{\mathrm{m}}=BA$

This expression of${\Phi }_{\mathrm{m}}$into equation to get localid="1648967434879" $\epsilon$by

$\epsilon =\frac{d\left(BA\right)}{dt}=B\frac{dA}{dt}$

The area of the sphere is $\pi r^2$, and as the volume expands, this means the radius of the sphere changes. So, we use the expression of $A$into equation and differentiate it for localid="1648967449782" $r$

$\epsilon =B\frac{d\left(\pi r^2\right)}{dt}=2\pi Br\left(\frac{dr}{dt}\right).$

Step: 2 Finding the value of radius:

Since the volume increases at $dV/dt=0.75\mathrm{L}/\mathrm{s}$a constant rate of, we can calculate the radius expansion rate$dr/dt$by multiplying by

$\begin{array}{l}V=\frac{4}{3}\pi r^3\\ \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\\ \frac{dr}{dt}=\frac{1}{4\pi r^2}\frac{dV}{dt}\end{array}$

The expression of $(dr/dt)$into equation to get

$\epsilon =2\pi Br\left(\frac{1}{4\pi r^2}\frac{dV}{dt}\right)=\left(\frac{B}{2r}\right)\frac{dV}{dt}$

We ignored the significance of. $r$. The starting volume is$V_o=2.5\mathrm{L}$, and it increases to over time$t=2\mathrm{s}$, $V$allowing us to get by.

localid="1648969318573" $V=V_o+\frac{dV}{dt}\left(dt\right)=2.5\mathrm{L}+(0.75\mathrm{L}/\mathrm{s})(2\mathrm{s})=4\mathrm{L}$

The final radius by

$\begin{array}{l}V=\frac{4}{3}\pi r^3\\ r=\sqrt[3]{\frac{3V}{4\pi }}\\ r=\sqrt[3]{\frac{3\left(4\times {10}^{-3}{\mathrm{m}}^3\right)}{4\pi }}\\ r=0.09\mathrm{m}.\end{array}$

Step: 3 Finding the value of current:

Putting the values of $B,r$and $\left(\raisebox{1ex}{$dV$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.\right)$into the equation,

$\begin{array}{l}\epsilon =\left(\frac{B}{2r}\right)\frac{dV}{dt}\\ \epsilon =\left(\frac{45\times {10}^{-3}\mathrm{T}}{2\left(0.09\mathrm{m}\right)}\right)\left(0.75\times {10}^{-3}{\mathrm{m}}^3/\mathrm{s}\right)\\ \epsilon =1.875\times {10}^{-4}\mathrm{V}.\end{array}$

To get current, we utilise ohm's law.

$$\begin{gathered} I=\frac{\epsilon }{R} \\ I=\frac{1.875\times {10}^{-4}\mathrm{V}}{2.5\mathrm{\Omega }} \\ I=7.5\times {10}^{-5}\mathrm{A}. \end{gathered}$$