Question

An infintely 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 ?

Answer 1

Force on AC:
Force on circular portions of the circuit i.e.,AC etc., due to the wire at the centre will be zero because magnetic field due to the central wire at these arcs will be tangential ($\theta ={180}^o$).
Force on CD:
Current in central wire is also i=10 A.Magnetic field at distance x due to central wire,
$B=\frac{{\mu }_o}{2\pi }\cdot \frac{i}{x}$\therefore$ Magnetic force on element dx due to this magnetic field
$dF=\left(i\right)(\frac{{\mu }_o}{2\pi }\cdot \frac{i}{x})\cdot dx=(\frac{{\mu }_o}{2\pi })i^2\frac{dx}{x}$
$(F=ilBsin{90}^o)$

Therefore, net force on CD is:
$F={\int }_{x=r_1}^{x=r_2}dF=\frac{{\mu }_o{i}^2}{2\pi }{\int }_{0.08}^{0.12}\frac{dx}{x}=\frac{{\mu }_o}{2\pi }{i}^2\ln \left(\frac{3}{2}\right)$

Substituing the values,
$F=(2\times {10}^{-7}){\left(10\right)}^2\ln \left(1.5\right)$
$F=8.1\times {10}^{-6}N$ (inwards)

Force on wire at the centre
Net magnetic field at the centre due to the circuit is in vertical direction and current in the wire in centre is also in vertical direction.Therefore,net force on the wire at the centre will be zero.$(\theta ={180}^o)$.Hence,
(i) Force acting on the wire at the centre is zero.
(ii) Force on arc AC=0.
(iii) Force on segment CD is $8.1\times {10}^{-6}N$(inwards)