Question

$A100$turn, $8.0cm$diameter coil is made of $0.50mm$diameter copper wire. A magnetic field is parallel to the axis of the coil. At what rate must $B$increase to induce a $2.0A$current in the coil?

Short answer

This is the speed of current in the coil$\frac{dB}{dt}=8.7\mathrm{T}/\mathrm{s}$

Step-by-step solution

Law of electricity (part a)

The frequency of on the spot is directly proportional to the potential drop and inversely proportional to the circuit resistance, per the law of electricity.

Explanations (part b)

We utilize Ohm's law to induce the induced current L across the loop, as indicated within the next equation.

Equation 1

$I_{\text{induced}}=\frac{\epsilon }{R}$

With $\epsilon$is that the portal's induced emf. The induced emf, as defined by Faraday's law, is that the change in magnetic flux inside the loop, and it is provided by equation within the format

Equation $2$

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

With ${\mathrm{\Phi }}_{\mathrm{m}}$is that the flux through to the loop, which is that the amount of force field that flows through a loop of area $A$, and therefore the flux through the loop, which is defined by

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

Let's plug this ${\mathrm{\Phi }}_{\mathrm{m}}$expression into equation ($2$) to urge$\epsilon$.

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

The loop's area is computed by

$A=\pi {\left(\frac{d}{2}\right)}^2=\pi {\left(\frac{0.08\mathrm{m}}{2}\right)}^2=5\times {10}^{-3}{\mathrm{m}}^2$

The wire's resistance $R$ is decided by its geometry and is given by
$R=\frac{\rho L}{A_w}$

Where $\rho$denotes the copper's resistivity, $L$denotes the wire's length, and $A_w$denotes the wire's area. The wire's area is computed by

$A_{\mathrm{w}}=\pi {\left(\frac{d_{\mathrm{w}}}{2}\right)}^2=\pi {\left(\frac{0.5\times {10}^{-3}\mathrm{m}}{2}\right)}^2=1.96\times {10}^{-7}{\mathrm{m}}^2$

The wire's size is set to the radius of the coil, which is capable the number of$N$ turns.
$L=\left(\pi d\right)N$

As a result, the wire's resistance is

Equation $3$
$R=\frac{\pi \rho dN}{A_{\mathrm{w}}}$

Find the speed of current

To get an expression for $dB/dt$, we plug the expression of $R$and into equation (Ohm's law).
$$\begin{gathered} I=\frac{\epsilon }{R} \\ I=\frac{\left(NA\frac{dB}{dt}\right)}{\left(\frac{\pi \rho dN}{A_w}\right)} \end{gathered}$$

$\frac{dB}{dt}=\frac{\pi \rho dIN}{A_{w}NA}$

Equation $4$

$\frac{dB}{dt}=\frac{\pi \rho dI}{A_{\mathrm{w}}A}$

Let, we take the rates for $\rho ,d,1,A$and $A_{\mathrm{w}}$into equation ($4$) to seekout$dB/dt$

$\frac{dB}{dt}=\frac{\pi \rho dI}{A_{\mathrm{w}}A}$

$=\frac{\pi \left(1.7\times {10}^{-8}\right)(0.08\mathrm{m})(2\mathrm{A})}{\left(1.96\times {10}^{-7}{\mathrm{m}}^2\right)\left(5\times {10}^{-3}{\mathrm{m}}^2\right)}$

$=8.7\mathrm{T}/\mathrm{s}$