Question

The figure shows two parallel loops of wire having a common axis. The smaller loop (radius r) is above the larger loop (radius R) by a distancex>>R. Consequently, the magnetic field due to the counterclockwise current i in the larger loop is nearly uniform throughout the smaller loop. Suppose that x is increasing at the constant rate$\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{v}$. (a)Find an expression for the magnetic flux through the area of the smaller loop as a function of x. (b)In the smaller loop, find an expression for the induced emf. (c)Find the direction of the induced current.

Short answer

a) Expression for the magnetic flux through the smaller loop is,$\varphi =\frac{{\mathrm{\pi \mu }}_0{\mathrm{ir}}^2{\mathrm{R}}^2}{2x^3}$

b) Expression for the induced emf for the smaller loop is,$\epsilon =\frac{3\mathrm{\pi }{\mu }_{0}i{r}^2{R}^{2}V}{2x^4}$

c) Direction of induced current in the smaller loop is counterclockwise.

Step-by-step solution

Step 1: Given

i) Radius of the smaller loop isr

ii) Radius of the larger loop isR

iii) Distance between the two loops,x>>R

Determining the concept

Write the formula of the magnetic field due to the current carrying coilon its own axis and use it to calculate the magnetic flux through the smaller loop. Using Faraday’s law, find the induced emf and the direction of the induced current in the smaller loop.

Faraday'slaw of electromagnetic inductionstates, Whenever a conductor is placed in a varying magnetic field, an electromotive force is induced in it.

Formulae are as follows:

$$\begin{gathered} B\left(Z\right)=\frac{{\mu }_{0}i{R}^2}{2{\left(R^2+Z^2\right)}^{{\displaystyle \frac{3}{2}}}} \\ \varphi =\overrightarrow{B}.\overrightarrow{A} \end{gathered}$$

Where,$\Phi$is magnetic flux, B is magnetic field, A is area, i is current, R is resistance, 𝜇₀is permeability, z is the distance of the point.

(a) Determining the expression for the magnetic flux through the smaller loop

The magnetic field due to the current carrying coil on its own axis is given by,

$B\left(z\right)=\frac{{\mu }_{0}i{R}^2}{2{\left(R^2+Z^2\right)}^{{\displaystyle \frac{3}{2}}}}$

Where,Ris the radius of the coil,is the distance of the point from the center of the coil to the point on the axis of coil.

Here,Z=x.

Hence, the magnetic field will be,

role="math" localid="1661847252817" $B\left(Z\right)=\frac{{\mu }_{0}i{R}^2}{2{\left(R^2+x^2\right)}^{{\displaystyle \frac{3}{2}}}}$

Let’s assume that the distance is much greater than the radius of the coil, i.e.x>>R .

Therefore,

$$\begin{gathered} B\left(x\right)=\frac{{\mu }_{0}i{R}^2}{2{\left(x^2\right)}^{{\displaystyle \frac{3}{2}}}} \\ B\left(x\right)=\frac{{\mu }_{0}i{R}^2}{2x^3} \end{gathered}$$

Here, consider that the +x axis is upwards, then in the vector notation, the magnetic field is,

$\overrightarrow{B}\left(x\right)=\frac{{\mu }_{0}i{R}^2}{2x^3}\stackrel{⏜}{i}$

Therefore, the magnetic flux through the smaller loop is given by,

Where,$A={\mathrm{\pi r}}^2$is the area of the smaller loop.

Thus,

$\varphi =\frac{\mathrm{\pi }{\mu }_{0}i{r}^2{R}^2}{2x^3}$

Hence, expression for the magnetic flux through the smaller loop is,$\varphi =\frac{\mathrm{\pi }{\mu }_{0}i{r}^2{R}^2}{2x^3}$ .

(b) Determining the expression for the induced emf for the smaller loop

Faraday’s law is given by,

$\epsilon =-\frac{d\varphi }{dt}$

Substituting the value of$\varphi$,

$\epsilon =-\frac{d}{dt}\left(\frac{{\mathrm{\pi \mu }}_0{\mathrm{ir}}^2{\mathrm{R}}^2}{2x^3}\right)$

Where, r are R the radii of the smaller and bigger loops respectively and are constants. Take the constants outside the derivative. So,

$$\begin{gathered} \epsilon =-\left(\frac{\mathrm{\pi }{\mu }_{0}i{r}^2{R}^2}{2}\right)\frac{d}{dt}\left(\frac{1}{x^3}\right) \\ \epsilon =-\left(\frac{\mathrm{\pi }{\mu }_{0}i{r}^2{R}^2}{2}\right)\left(-\frac{3}{x^4}\right)\frac{dx}{dt} \end{gathered}$$

Since,$\frac{dx}{dt}=V$,

Therefore,

$\epsilon =\frac{3{\mathrm{\pi \mu }}_0{\mathrm{ir}}^2{\mathrm{R}}^2\mathrm{V}}{2x^4}$

Hence, expression for the induced emf for the smaller loop is,$\epsilon =\frac{3{\mathrm{\pi \mu }}_0{\mathrm{ir}}^2{\mathrm{R}}^2\mathrm{V}}{2x^4}$.

(c) Determining the direction of the induced current in the smaller loop

As the smaller loop moves upward, the flux through the loop decreases. The induced current will be directed so as to produce a magnetic field which is upward. So, the induced current will be counterclockwise, in the same direction as the current in the larger loop.

Hence, the direction of induced current in the smaller loop is counterclockwise.

Therefore, use the formula for the magnetic field due to the current carrying coil to calculate the expression for the magnetic flux and Faraday’s law to calculate the emf induced and the direction of the current.