Question

An $8.0\mathrm{cm}\times 8.0\mathrm{cm}$square loop is halfway into a magnetic CALC field perpendicular to the plane of the loop. The loop's mass is and its resistance is $0.010\mathrm{\Omega }$. A switch is closed at $\mathrm{t}=0\mathrm{s}$, causing the magnetic field to increase from$0$to$1.0\mathrm{T}$in$0.010\mathrm{s}$ .

a. What is the induced current in the square loop?

b. With what speed is the loop "kicked" away from the magnetic field?

Hint: What is the impulse on the loop?

Short answer

a. Induced current in the square loop is$32\mathrm{A}$.

b. Speed of loop is$0.64\mathrm{m}/\mathrm{s}$.

Step-by-step solution

Formula 

Ohm's law as ,

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

In the loop's induced emf is localid="1648921569096" $\mathrm{\epsilon }$.

The induced emf, according to Faraday's law, is the change in magnetic flux inside the loop.

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

Where localid="1648811856380" ${\mathrm{\Phi }}_{\mathrm{m}}$is the flux through the loop.

Magnetic field that flows through a loop of arealocalid="1648921574407" $\mathrm{A}$is,

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

So,

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

From that,

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

Calculation for induced current in the square loop (part a)

(a).

Because the magnetic field ,passes through half of the loop, the total flux area is computed by

$\mathrm{A}=\frac{1}{2}(0.08\mathrm{m})(0.08\mathrm{m})=32\times {10}^{-4}{\mathrm{m}}^2$

The magnetic field changes fromlocalid="1648921523389" $0\mathrm{T}$tolocalid="1648921529152" $1\mathrm{T}$in time localid="1648921534519" $\mathrm{dt}=0.010\mathrm{s}$.

The shift in magnetic field as a result of,

$\frac{\mathrm{dB}}{\mathrm{dt}}=\frac{1\mathrm{T}-0\mathrm{T}}{0.010\mathrm{s}}=100\mathrm{T}/\mathrm{s}$

So,

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

$=\left(\frac{32\times {10}^{-4}{\mathrm{m}}^2}{0.010\mathrm{\Omega }}\right)$

$=32\mathrm{A}$

Calculation for speed (part b)

(b).

The impulse is the amount of energy needed to change the loop's motion.

$\mathrm{J}=\mathrm{m}\mathrm{d}\mathrm{v}$

$\mathrm{J}=\mathrm{F}\mathrm{d}\mathrm{t}$

Equating both,

$\mathrm{F}\mathrm{d}\mathrm{t}=\mathrm{m}\mathrm{d}\mathrm{v}$

$\frac{1}{2}\mathrm{IlBdt}=\mathrm{mdv}$

$\frac{1}{2}\frac{\mathrm{A}}{\mathrm{R}}\frac{\mathrm{dB}}{\mathrm{dt}}\mathrm{lBdt}=\mathrm{mdv}$

$\mathrm{dv}=\frac{\mathrm{Al}}{2\mathrm{Rm}}\mathrm{BdB}$

Integrate localid="1648921549494" $\mathrm{dv}$,

$\mathrm{v}=\frac{\mathrm{Al}}{2\mathrm{Rm}}\int \mathrm{BdB}=\left(\frac{\mathrm{Al}}{2\mathrm{Rm}}\right)\frac{{\mathrm{B}}^2}{2}=\frac{{\mathrm{AlB}}^2}{4\mathrm{Rm}}$$=\frac{\left(32\times {10}^{-4}{\mathrm{m}}^2\right)(0.08\mathrm{m})(1\mathrm{T}{)}^2}{4(0.010\mathrm{\Omega })\left(10\times {10}^{-3}\mathrm{kg}\right)}$

$=0.64\mathrm{m}/\mathrm{s}$