Question

A $20\mathrm{cm}\times 20\mathrm{cm}$square loop has a resistance of $0.10\mathrm{\Omega }$. A CALC magnetic field perpendicular to the loop is $\mathrm{B}=4\mathrm{t}-2{\mathrm{t}}^2$, where $\mathrm{B}$is in Tesla and$\mathrm{t}$is in seconds. What is the current in the loop at $\mathrm{t}=0.0\mathrm{s},\mathrm{t}=1.0\mathrm{s}$,and$\mathrm{t}=2.0\mathrm{s}?$

Short answer

Current in the loop at ${\mathrm{I}}_{0\mathrm{s}}$is $1.6\mathrm{A}$.

Current in the loop at ${\mathrm{I}}_{1\mathrm{s}}$is $0\mathrm{A}$.

Current in the loop at ${\mathrm{I}}_{2\mathrm{s}}$is $-1.6\mathrm{A}$.

Step-by-step solution

Formula for electric flux inside the loop

The magnetic field is expressed in terms of time in the form

$\mathrm{B}=4\mathrm{t}-2{\mathrm{t}}^2$

The loop's current state is,

${\mathrm{I}}_{\text{induced}}=\frac{\mathrm{\epsilon }}{\mathrm{R}}$

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

The electric flow as a function of area is,

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

So,

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

Calculation for current in the loop att=0.0 s

Current that has been induced is,

${\mathrm{I}}_{\text{induced}}=\left(\frac{\mathrm{NA}}{\mathrm{R}}\right)\frac{\mathrm{dB}}{\mathrm{dt}}$

$=\left(\frac{\mathrm{NA}}{\mathrm{R}}\right)\frac{\mathrm{d}}{\mathrm{dt}}\left(4\mathrm{t}-2{\mathrm{t}}^2\right)$

$=\left(\frac{\mathrm{NA}}{\mathrm{R}}\right)(4-4\mathrm{t})$

The area of the loop is calculated using

$\mathrm{A}=0.20\mathrm{m}\times 0.20\mathrm{m}=4\times {10}^{-2}{\mathrm{m}}^2$

At $\mathrm{t}=0.0\mathrm{s}$:

Induced current is,

${\mathrm{I}}_{0\mathrm{s}}=\left(\frac{\mathrm{NA}}{\mathrm{R}}\right)(4-4\mathrm{t})$

$=\left(\frac{\left(1\right)\left(4\times {10}^{-2}{\mathrm{m}}^2\right)}{0.10\mathrm{\Omega }}\right)(4-4(0\mathrm{s}\left)\right)$

$=1.6\mathrm{A}$

Calculation for current in the loop at t=1.0 s and t=2.0 s

At $\mathrm{t}=1.0\mathrm{s}$:

Induced current is,

${\mathrm{I}}_{1\mathrm{s}}=\left(\frac{\mathrm{NA}}{\mathrm{R}}\right)(4-4\mathrm{t})$

$=\left(\frac{\left(1\right)\left(4\times {10}^{-2}{\mathrm{m}}^2\right)}{0.10\mathrm{\Omega }}\right)(4-4(1\mathrm{s}\left)\right)$

$=0\mathrm{A}$

At $\mathrm{t}=2.0\mathrm{s}$:

Induced current is,

${\mathrm{I}}_{2\mathrm{s}}=\left(\frac{\mathrm{NA}}{\mathrm{R}}\right)(4-4\mathrm{t})$

$=\left(\frac{\left(1\right)\left(4\times {10}^{-2}{\mathrm{m}}^2\right)}{0.10\mathrm{\Omega }}\right)(4-4(2\mathrm{s}\left)\right)$

$-1.6\mathrm{A}$$=\left(\frac{\left(1\right)\left(4\times {10}^{-2}{\mathrm{m}}^2\right)}{0.10\mathrm{\Omega }}\right)(4-4(1\mathrm{s}\left)\right)$