Question

A current of $ 10 A$flows around a closed path in a circuit which is in the horizontal plane as shown in the figure. The circuit consists of eight alternating areas of radii$ {r}_{1}=0.08m and {r}_{2}=0.12m$. Each arc subtends the same angle at the centre.
(a) Find the magnetic field produced by this circuit at the centre.
(b) An infinitely long straight wire carrying a current of$ 10 A$is passing through the centre of the above circuit vertically with the direction of the current being into the plane of the circuit. What is the force acting on the wire at the centre due to the current in the circuit? What is the force acting on the arc AC and the straight segment CD due to the current at the centre?

BBYJU'S

Answer 1

Step 1: Given data

Radii of smaller arcs, $r_1=0.08m$

Radii of larger arcs, $r_2=0.12m$

Current flowing through the given circuit, $i=10A$

Magnetic permeability, ${\mu }_0=4\pi \times {10}^{-7}$

Step 2: Calculation of the magnetic field produced at the centre of the circuit

The magnetic field at the centre of the given circuit will be due to the current flowing in the smaller and larger arcs.

The smaller arcs can be treated as a single semicircle of radius ($r_1$) and the larger arcs can be treated as a single semicircle of radius ($r_2$)

Since the current flowing through these semicircles is in opposite direction, therefore the magnetic field produced by them at the centre of the circuit will be equal to the summation of the individual magnetic field produced by them.

$B_{net}=B_1+B_2$ ……………(a)

Where ($B_{net}$) is the net magnetic field at the centre of the circuit.

$B_1$is the magnetic field due to the smaller semicircle

$B_2$is the magnetic field due to the larger semicircle

We know that the magnetic field ($B$) at the centre of a semicircular wire of radius ($r$) due to the current ($i$) flowing through it given by

$B=\frac{{\mu }_{0}i}{4r}$……..(b)

Using equation (b) in equation (a) and using the corresponding values of radius and the current, we get

$B_{net}=\frac{{\mu }_{0}i}{4r_1}+\frac{{\mu }_{0}i}{4r_2}$……….(c)

Substituting the given values in equation (c), we get

$B_{net}=\frac{{\mu }_{0}i}{4}\left[\frac{1}{r_1}+\frac{1}{r_2}\right]$

Substituting the given values in the above equation, we get

$B_{net}=\frac{4\pi \times {10}^{-7}\times 10}{4}\left[\frac{1}{0.08}+\frac{1}{0.12}\right]$

$B_{net}=6.54\times {10}^{-5}T$ (outwards)

On pointing our right thumb in the direction of the current, our fingers curl in an outward direction which gives the direction of the magnetic field at the centre of the circuit due to the current flowing in the two semicircle portions of the circuit.

Step 4: Calculation of the magnetic force acting on the wire at the centre due to the current flowing through the circuit

We know that the magnetic force ($F$) acting on a current-carrying conductor of length ($L$) placed at an angle of ($\theta$) in a magnetic field of strength ($B$) is given by

$F=iLB\sin \theta$

In the given case, $\theta ={180}^0\Rightarrow \sin {180}^0=0\Rightarrow F=0$

Hence, the force acting on the wire is zero

Step 5: Calculation of the force acting on the arc AC

The magnetic field ( $B_{AC}$) due to the current flowing in the arc AC acts in the circumferential direction and is given by

$B_{AC}=\frac{{\mu }_{o}i}{2\pi r}$

$i$ denotes the current flowing through the arc AC

$r$ denotes the radial distance of any point on the arc AC from the centre

The force ($F_{AC}$) on the arc “AC” of length ($L$) due to the magnetic field ($B$) is given by

$F_{AC}=iLB\sin \theta$

$\theta$ is the angle between the magnetic field and the electric current.

The direction of the magnetic field on any small segment of arc AC will be tangential because the magnetic field is opposite to the direction of the electric current.

In the given case, $\theta ={180}^0\Rightarrow \sin {180}^0=0\Rightarrow F=0$

Hence, the force acting on the arc AC is equal to zero.

Step 6: Calculation of the force acting on the segment CD

Consider a small length element ($dx$) on the line segment CD as shown in the diagram above

The small force($dF$) acting on a small length element ($dx$) of the segment CD carrying a current ($i$) due to the magnetic field ($B$) is given by

$dF=idxB$ ………….(a)

The magnetic field ($B$) on the small length element ($dx$) located at a radial distance of ($x$) from the centre is given by

$B=\frac{{\mu }_{0}i}{2\pi x}$…………(b)

Substituting equation (b) in equation (a), we get

$dF=i\times dx\times \frac{{\mu }_{0}i}{2\pi x}$

Substituting the given value of current ( $i=10A$) in the above equation, we get

$$\begin{gathered} dF=10\times dx\times \frac{{\mu }_0\times 10}{2\pi x} \\ dF=\frac{50{\mu }_{0}i}{\pi }\times \frac{dx}{x} \end{gathered}$$

The total force on the segment CD can be calculated by integrating the above equation between the limits ($r_1$) and ($r_2$)

$$\begin{gathered} \int dF={\int }_{r_1}^{r_2}\frac{50{\mu }_0}{\pi }\times \frac{dx}{x} \\ F=\frac{50{\mu }_0}{\pi }{\int }_{r_1}^{r_2}\frac{dx}{x} \\ F=\frac{50{\mu }_0}{\pi }{\left[{\log }_{e}x\right]}_{r_1}^{r_2} \\ F=\frac{50{\mu }_0}{\pi }\left[{\log }_e{r}_2-{\log }_e{r}_1\right] \end{gathered}$$

$F=\frac{50{\mu }_0}{\pi }\times {\log }_e\left[\frac{r_2}{r_1}\right]$ ……………….(w)

Substituting the given values in equation (w), we get

$$\begin{gathered} F=\frac{50\times 4\pi \times {10}^{-7}}{\pi }\times {\log }_e\left[\frac{0.12}{0.08}\right] \\ F=8.1\times {10}^{-6}N \end{gathered}$$

Hence, the force acting on the segment CD is equal to $8.1\times {10}^{-6}N$