Question

FIGURE shows a $4.0$-cm-diameter loop with resistance $0.10\Omega$around a $2.0$-cm-diameter solenoid. The solenoid is $10$ cm long, has $100$turns, and carries the current shown in the graph. A positive current is $cw$when seen from the left. Find the current in the loop at$\left(a\right)t=0.5s,\left(b\right)t=1.5s,and\left(c\right)t=2.5s.$

Short answer

(a) By assign,$t=0.5\to I_{\text{induced}}=0\mathrm{A}$

(b) By assign,$t=1.5\to I_{\text{induced}}=12.56\times {10}^{-5}\mathrm{A}$

(c) By assign,$t=2.5\to I_{\text{induced}}=0\mathrm{A}$

Step-by-step solution

Introduction

Given values:

$R=0.1\Omega$

$n=1000$

$r=R^{\text{'}}=2\times {10}^{-2}\mathrm{m}$

$r=1\times {10}^{-2}$

$B_s={\mu }_{o}nI$

Use next equations,

$E\left(2\pi R\right)=\pi r^2\frac{dB}{dt}$

$E\left(2\pi R^{\text{'}}\right)=\pi r^2{\mu }_{o}n\frac{dI}{dt}$

$E=\frac{\pi r^2}{2\pi R^{\text{'}}}{\mu }_{o}n\frac{dI}{dt}$

$E=\frac{r^2{\mu }_{o}n}{2R^{\text{'}}}\frac{dI}{dt}$

$I_{\text{induced}}=\frac{{\left({10}^{-2}\right)}^2}{2\times 2\times {10}^{-2}}\times \left(4\pi \times {10}^{-7}\right)\times 1000\frac{dI}{dt}=3.14\times {10}^{-6}\frac{dI}{dt}$

Find Iinduced

a.) $I_{\text{induced}}$at $t=0.5$:

$I_{\text{induced}}=3.14\times {10}^{-6}\times 0$

$I_{\text{induced}}=0\mathrm{A}\to \left|\frac{dI}{dt}=0\right|$

b.) $I_{\text{induced}}$at $t=1.5$:

$I_{\text{induced}}=3.14\times {10}^{-6}\times 40$

$I_{\text{induced}}=12.56\times {10}^{-5}\mathrm{A}\to \left|\frac{dI}{dt}=40\right|$

c.) $I_{\text{induced}}$at $t=2.5$:

$I_{\text{induced}}=0\mathrm{A}\to \left|\frac{dI}{dt}=0\right|$

Explain with graph