Question

Figure shows two parallel wires separated by a distance of 4.0 cm and carrying equal currents of 10 A along opposite directions. Find the magnitude of the magnetic field B at the points A₁, A₂, A₃. 

Answer 1

For point A₁,
Magnitude of current in wires, I = 10 A 
Separation of point A₁ from the wire on the left side, d = 2 cm
Separation of point A₁ from the wire on the right side, d' = 6 cm 

In the figure
Red and blue arrow denotes the direction of magnetic field due to the wire marked as red and blue respectively.
P (marked red) denotes the wire carrying current in a plane going into the paper.
Q (marked blue) denotes the wire carrying current in a plane coming out of the paper.
Also from the figure, we can see that

\[P A_4    =   Q A_4   \] 

\[\angle   A_4  A_3 P   =   \angle A_4  A_3 Q   =   90^\circ \] 

\[ \Rightarrow \angle   A_4 P A_3    =   \angle   A_4 Q A_3    =   45^\circ \] 

\[ \Rightarrow \angle  P A_4  A_3    =   \angle  Q A_4  A_3    =   45^\circ \] 

\[ \Rightarrow \angle  P A_4 Q   =   90^\circ \]

The magnetic field at A₁ due to current in the wires is given by

\[B = \frac{\mu_0 I}{2\pi d} - \frac{\mu_0 I}{2\pi d'}\]    ...(1)

\[\Rightarrow B =   \frac{2 \times {10}^{- 7} \times 10}{2 \times {10}^{- 2}} - \frac{2 \times {10}^{- 7} \times 10}{6 \times {10}^{- 2}}\] 

\[                 =   \left( 1 - \frac{1}{3} \right) \times  {10}^{- 4} \] 

\[                 =   0 . 67 \times  {10}^{- 4}   T\]

Similarly, we get the magnetic field at A₂ using eq. (1).

\[B =   \frac{2 \times {10}^{- 7} \times 10}{1 \times {10}^{- 2}} + \frac{2 \times {10}^{- 7} \times 10}{3 \times {10}^{- 2}}\] 

\[       = \frac{8}{3} \times  {10}^{- 4}   T\] 

\[       = 2 . 67 \times  {10}^{- 4}   T\]

Now,
Magnetic field at A₃:

\[B   =   \frac{2 \times {10}^{- 7} \times 10}{2 \times {10}^{- 2}} + \frac{2 \times {10}^{- 7} \times 10}{2 \times {10}^{- 2}}\] 

\[       =   2 \times  {10}^{- 4}   T\]

Magnetic field at A₄:
Separation of point A₄ from the wire on the left side, d = \[\sqrt{2^2 + 2^2}   = 2\sqrt{2} \] cm 

Separation of point A₄ from the wire on the right side, d' =\[\sqrt{2^2 + 2^2}   = 2\sqrt{2}  \] cm 

Thus, the magnetic field at A₄ due to current in the wires is given by

 

\[B   =   \sqrt{\left( \frac{2 \times {10}^{- 7} \times 10}{2\sqrt{2} \times {10}^{- 2}} \right)^2 + \left( \frac{2 \times {10}^{- 7} \times 10}{2\sqrt{2} \times {10}^{- 2}} \right)^2}\] 

\[       = 1 \times  {10}^{- 4}   T\]