# 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 arcs of radii r1 = 0.08 m and r1 = 0. 12 m. Each are subtends the same angle at the centre.  Find the magnetic field produced by this circuit at the center.  (IIT JEE 2001 – 10 MARKS)  An infinitely long straight wire carrying a current of 10 A is passing through the center 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 center 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 center ?

Deepak Patra
10 years ago
Hello Student,
For finding the magnetic field produced by this circuit at the centre we can consider it to contain two semicircles of radii, r1 = 0.08 m and r2 = 0.12 m. Since current is flowing in the same direction, the magnetic field created by circular arcs will be in the same direction and therefore will be added.
∴ B1 = μ0i / 4r1 and B2 = μ­0i / 4r2 ∴ B = μ0i /4 [1 / r1 + 1 / r2]
∴ B = (6 . 54 x 10-5) T Directed outwards
(Right hand thumb rule )
(b) Force acting on a current carrying conductor placed in a magnetic field is given by
1. For force acting on the wire at the centre In this case θ = 180°
∴ F = 0
1. On are AC to current at the centre
| | on AC will be B = μ0I / 2πr1
The direction of this magnetic field on any small segment of AC will be tangential
∴ θ = 180° ⇒ F = 0
1. On segment CD.
Force on a small segment dx distant r from O
dF = I dx B
= 10 x dx x μ0I / 2πx = 5μ0 I / π dx / x
∴ F = 5 μ­0 I / π loge r2 / r1 = 5μ0 x 10 / π loge (0.12 / 0.08)
= 8.1 x 10-6 N
directed downwards (By Fleming left hand rule).
Thanks
Deepak patra