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 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?

Answer 1

(a)

To calculate the magnetic field at the center of the center, let

us consider two semi circles of radii $r_1$= 0.8m and $r_2$= 0.12m.

Since, current is flowing in the same direction. The magnetic field

created by circular arcs will be in same direction.

$\therefore B_1=\frac{{\mu }_{o}i}{4r_1}$

And $B_2=\frac{{\mu }_{o}I}{4[\frac{1}{r_1}+\frac{1}{r_2}]}$

Therefore $B=6.54\times {10}^{-5}T$ outward (By right hand thumb rule)

(b)

Force acting on a current carrying conductor placed in a magnetic

field is, $\overrightarrow{F}=I(l\times B)=BIlsin\theta$

(1) Force acting on the wire at the center, $\theta ={180}^0$,

F=0

On segment CD, $dF=Id\times B=10\times dx\times \frac{{\mu }_{o}I}{2\pi x}$

$\therefore F=\frac{5{\mu }_{o}I}{\pi {\int }_{r_2}^{r_2}\frac{dx}{x}}$

$\therefore F=\frac{5{\mu }_{o}I}{\pi log\left(\frac{0.12}{0.08}\right)}$

= $8.1\times {10}^{-6}N$

The direction of force is downward.