(a) Derive the expression for the force acting per unit length between two long straight parallel current carrying conductors. Hence define one ampere.

(b) Two long parallel straight conductors are placed 12 cm apart in air. They carry equal currents of 3 A each. Find the magnitude and direction of the magnetic field at a point midway between them (drawing a figure) when the currents in them flow in opposite directions.

 

 

 
 
 
 
 

Answers (1)

a)

Consider a small length L of the long straight conductor

B_{1}=\frac{\mu_{o}I_{1}}{2\pi d}

B_{2}=\frac{\mu_{o}I_{2}}{2\pi d}

F_{12}=I_{2}B_{1}L

F_{21}=I_{1}B_{2}L

F = F_{12}=F_{21}=\frac{\mu_{o}I_1I_{2}L}{2\pi d}

\frac{F}{L}=\frac{\mu_{o}I_{1}I_{2}}{2\pi d}.

Ampere is that value of steady-state current, which when maintained in each of the very two long, straight, parallel conductors of negligible cross-section and placed 1m apart in a vacuum produces 2\times 10^{-7}\; N/m force on each conductor. 

b) 

\\\vec{B}=\vec{B_1}+\vec{B_2}\\\\=\frac{\mu_0 I_1}{2\pi r_1}+\frac{\mu_0 I_2}{2\pi r_2}\\here \ I_1=I_2\\r_1=r_2\\therefore\\B=2\frac{\mu_0 I_1}{2\pi r_1}=\frac{4\pi\times10^{-7}\times3}{\pi\times6\times10^{-2}}=2\times10^{-5}T

The direction of B at the midpoint is perpendicular to the plane containing the two conductors and pointing downwards.

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