FIGURE$\mathrm{P}30.48$shows two $20$-turn coils tightly wrapped on the same$2.0-\mathrm{cm}$-diameter cylinder with $1.0-\mathrm{mm}$-diameter wire. The current through coil $1$is shown in the graph. Determine the current in coil $2$at (a) $\mathrm{t}=0.05\mathrm{s}$and (b) . A positive current is $\mathrm{t}=0.25\mathrm{s}$into the page at the top of a loop. Assume that the magnetic field of coil localid="1648920663723" $1$passes entirely through coil localid="1648920667724" $2$.

Current,

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

Magnetic field for coillocalid="1648920724674" $1$is,

$\mathrm{B}=\frac{{\mathrm{\mu }}_{\mathrm{o}}{\mathrm{N}}_1\mathrm{I}}{\mathrm{l}}$

Wherelocalid="1648920735299" $\mathrm{l}$is the length of coil.

$\mathrm{l}={\mathrm{N}}_1{\mathrm{d}}_{\text{wire}}$

$=\left(20\right)(1\mathrm{mm})=20\mathrm{mm}$

Wherelocalid="1648920755478" ${\mathrm{d}}_{\text{wire}}$is the diameter of the wire.

$\mathrm{\epsilon }=\frac{{\mathrm{d\Phi }}_{\mathrm{m}}}{\mathrm{dt}}=\frac{\mathrm{d}\left({\mathrm{BAN}}_2\right)}{\mathrm{dt}}={\mathrm{N}}_2\mathrm{A}\frac{\mathrm{dB}}{\mathrm{dt}}$

The coil's surface area is,

$\mathrm{A}=\mathrm{\pi }{\left(\frac{{\mathrm{d}}_2}{2}\right)}^2$

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

(a).

Substitute$\mathrm{\epsilon }$and$\mathrm{B}$values,

${\mathrm{I}}_{\text{induced}}=\left(\frac{{\mathrm{N}}_2\mathrm{A}}{\mathrm{R}}\right)\frac{\mathrm{dB}}{\mathrm{dt}}$

$=\left(\frac{{\mathrm{N}}_2\mathrm{A}}{\mathrm{R}}\right)\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{{\mathrm{\mu }}_{\mathrm{o}}{\mathrm{N}}_1\mathrm{I}}{\mathrm{l}}\right)$

$=\frac{{\mathrm{A\mu }}_{\mathrm{o}}{\mathrm{N}}_1{\mathrm{N}}_2}{\mathrm{Rl}}\frac{\mathrm{dI}}{\mathrm{dt}}$

Atlocalid="1648920874161" $\mathrm{t}=0.05\mathrm{s}$,

Because the current is continuous, there is no change in current over time.

$\frac{\mathrm{dI}}{\mathrm{dt}}=0$

That the loop's induced current will be zero.

${\mathrm{I}}_{\text{induced}}=\frac{{\mathrm{A\mu }}_{\mathrm{o}}{\mathrm{N}}_1{\mathrm{N}}_2}{\mathrm{Rl}}\frac{\mathrm{dI}}{\mathrm{dt}}$

$\frac{{\mathrm{A\mu }}_{\mathrm{o}}{\mathrm{N}}_1{\mathrm{N}}_2}{\mathrm{Rl}}(0\mathrm{A}/\mathrm{s})=0\mathrm{A}$

(b).

At$\mathrm{t}=0.25\mathrm{s}$, the current changes.

So, between the two points, $0.1\mathrm{s}$and $0.3\mathrm{s}$the slope is

$\frac{\mathrm{dI}}{\mathrm{dt}}=\frac{2\mathrm{A}-(-2\mathrm{A})}{0.3\mathrm{s}-0.1\mathrm{s}}=20\mathrm{A}/\mathrm{s}$

All values should be substituted,

${\mathrm{I}}_{\text{induced}}=\frac{{\mathrm{A\mu }}_{\mathrm{o}}{\mathrm{N}}_1{\mathrm{N}}_2}{\mathrm{Rl}}\frac{\mathrm{dI}}{\mathrm{dt}}$

$=\frac{\left(3.14\times {10}^{-4}{\mathrm{m}}^2\right)\left(4\mathrm{\pi }\times {10}^{-7}\mathrm{T}·\mathrm{m}/\mathrm{A}\right)\left(20\right)\left(20\right)}{\left(2\mathrm{\Omega }\right)\left(20\times {10}^{-3}\mathrm{m}\right)}(20\mathrm{A}/\mathrm{s})$

$=79\times {10}^{-6}\mathrm{A}$$=79\mathrm{\mu A}$