Question

A very long straight wire (essentially infinite in length) carries a current of $\mathbf{6}$ampere (Figure 22.60). The wire passes through the center of a circular metal ring of radius $\mathbf{2}\mathbf{}\mathit{c}\mathit{m}$and resistance $\mathbf{2}\mathit{\Omega }$that is perpendicular to the wire. If the current in the wire increases at a rate of $\mathbf{0}\mathbf{.}\mathbf{25}\mathbf{}\mathit{A}\mathbf{/}\mathit{s}$, what is the current in the ring? Explain carefully.

Short answer

The induced current in the ring is $3.14\times {10}^{-9}\text{A}$.

Step-by-step solution

A concept:

An electric current occurs when a second conductor (a substance that carries electricity) is placed in an area where there is already an electric current.

A given data:

The radius, $r=4\text{cm}=0.04\text{m}$

The resistance, $R=2\Omega$

The current in the wire increases at a rate of$\frac{dI}{dt}=0.25A/s$

The induced current in the ring:

You know that for a long current carrying wire, there is a magnetic field at a distance $r$ from the wire.

$\overrightarrow{B}=\frac{{\mu }_{0}i}{2\pi r}$

But direction of this magnetic field is always perpendicular to the wire.

The magnitude induced emf in the ring is,

$$\begin{gathered} \epsilon =\frac{d\varphi }{dt} \\ =A\frac{{\mu }_0}{2\pi R}\frac{dI}{dt} \\ =\left(\pi r^2\right)\frac{{\mu }_0}{2\pi R}\frac{dI}{dt} \end{gathered}$$

Here, the permeability of free space is ${\mu }_0$having a value $4\pi \times {10}^{-7}H/m$.

Substitute known values in the above equation.

$$\begin{gathered} \epsilon =3.14\times {\left(0.04\text{m}\right)}^2\times \frac{4\pi \times {10}^{-7}H/m}{2\times 3.14\times 2\Omega }\times 0.25A/s \\ =6.28\times {10}^{-9}\text{V} \end{gathered}$$

The induced current in the ring is,

$$\begin{gathered} I=\frac{\epsilon }{R} \\ =\frac{6.28\times {10}^{-9}\text{V}}{2\Omega } \\ =3.14\times {10}^{-9}\text{A} \end{gathered}$$

Hence, the induced current in the ring is $3.14\times {10}^{-9}\text{A}$.