Question

Consider two long straight current

Consider two long, straight, current-carrying wire as shown in the Fig.Q3 One wire carries a current of 6.2 A in the positive y direction; the other wire carries a current of 4.5 A in the positive x direction. 

 (a) At which of the two points, A or B, do you expect the magnitude of the net magnetic field to be greater?  Explain. 

 (b) Find the magnitude and direction of the net magnetic field at points A and B.

Answer 1

Given data:

\(I_{1}=6.2 \mathrm{~A}\)

\(I_{2}=4.5 \mathrm{~A}\)

\(r_{1}=20 \times 10^{-2} \mathrm{~m}\)

\(r_{2}=16 \times 10^{-2} \mathrm{~m}\)

(a)

Applying Ampere's Right hand rule at the point \(\mathrm{A}\), The magnetic fiel d produced by the wire carrying \(6.2\) A current in the positive \(\mathrm{y}\) -direction is out of the page and let it be \(B_{1}\) The magnetic field produced by the wire carrying \(4.5 \mathrm{~A}\) current in the positive \(\mathrm{x}\) -direction is out of the page and let it be \(B_{2}\) The total magnetic field at the point \(\mathrm{A}\) is, \(B_{A}=B_{1}+B_{2}\) Applying Ampere's Right hand rule at the point \(B\), The magnetic fiel produced by the wire carrying \(6.2 \mathrm{~A}\) current in the positive \(y\) -direction is into of the page and let it be \(B_{1}\). The magnetic field produced by the wire carrying \(4.5\) A current in the positive \(\mathrm{x}\) -direction is out of the page and let it be \(B_{2}\) \(B_{B}=B_{2}-B_{1}\)

Since at the point \(A\), the magnetic field due to two wires is along the same direction and it adds up to produce a net magnetic field of greater strength. While the magnetic field produced due to two wires at point \(\mathrm{B}\) is opposite in direction which results in lesser net magnetic field

The net magnetic field at A has greater strength

(b)

\(B_{1}=\frac{\mu_{0} I_{1}}{2 \pi r_{1}}=\frac{\left(4 \pi \times 10^{-7}\right)(6.2)}{2 \pi\left(20 \times 10^{-2}\right)}=6.2 \times 10^{-6} \mathrm{~T}\)

\(B_{2}=\frac{\mu_{0} I_{2}}{2 \pi r_{2}}=\frac{\left(4 \pi \times 10^{-7}\right)(4.5)}{2 \pi\left(16 \times 10^{-2}\right)}=5.625 \times 10^{-5}\)

\(B_{A}=B_{1}+B_{2}=6.2 \times 10^{-6}+5.625 \times 10^{-6}=11.825 \times 10^{-6} \mathrm{~T}\)

\(B_{A}=11.825 \times 10^{-6} \mathrm{~T}\) and out of the page

\(B_{B}=-B_{1}+B_{2}=-6.2 \times 10^{-6}+5.625 \times 10^{-6}=-0.575 \times 10^{-6} \mathrm{~T}\)

\(B_{B}=0.575 \times 10^{-6} \mathrm{~T}\) into the page.