Question

Two long parallel straight wires P and Q separated by a distance 10cm in air carry currents of A and 2A respectively in opposite direction. The resultant magnetic field due to currents in these wires will be zero at

Answer 1

10 cm from wire P on the side away from wire Q

Answer 2

The magnetic field ($B$) due to a long straight wire carrying current ($I$) at a distance ($r$) is given by:

$B = \frac{\mu_0 I}{2 \pi r}$

Let the two wires be P and Q, separated by a distance $d = 10$ cm.
Current in wire P, $I_P = 1$ A.
Current in wire Q, $I_Q = 2$ A.
The currents are in opposite directions.

Let's analyze the direction of the magnetic field in different regions using the right-hand thumb rule. Assume current in P is out of the page and current in Q is into the page.

1. Region between the wires (0 < x < 10 cm):

   • For wire P (current out of page), the magnetic field to its right (between P and Q) is directed upwards.
   • For wire Q (current into page), the magnetic field to its left (between P and Q) is also directed upwards.

   Since the magnetic fields due to both wires are in the same direction in this region, they will add up. Therefore, the resultant magnetic field cannot be zero between the wires.

2. Region outside the wires:
   For the resultant magnetic field to be zero, the individual magnetic fields must be equal in magnitude and opposite in direction. This occurs outside the wires, closer to the wire carrying the smaller current. Since $I_P = 1$ A is smaller than $I_Q = 2$ A, the point where the field is zero will be closer to wire P, on the side away from Q.

   Let's place wire P at $x=0$ and wire Q at $x=10$ cm. The point where the magnetic field is zero will be at $x < 0$.
   Let the distance of this point from wire P be $r_P = |x|$.
   The distance of this point from wire Q will be $r_Q = d + |x| = 10 + |x|$.

   In this region ($x < 0$):

   • For wire P (current out of page), the magnetic field to its left is directed downwards.
   • For wire Q (current into page), the magnetic field to its left is directed upwards.

   The fields are in opposite directions, so they can cancel out.

   For the resultant magnetic field to be zero, the magnitudes of $B_P$ and $B_Q$ must be equal:

   $B_P = B_Q$

   $\frac{\mu_0 I_P}{2 \pi r_P} = \frac{\mu_0 I_Q}{2 \pi r_Q}$

   $\frac{I_P}{r_P} = \frac{I_Q}{r_Q}$

   Substituting the values:

   $\frac{1 \text{ A}}{r_P} = \frac{2 \text{ A}}{10 \text{ cm} + r_P}$

   $1 \times (10 \text{ cm} + r_P) = 2 \times r_P$

   $10 \text{ cm} + r_P = 2 r_P$

   $10 \text{ cm} = 2 r_P - r_P$

   $r_P = 10 \text{ cm}$

   So, the point where the magnetic field is zero is 10 cm from wire P. Since we established it's on the side away from Q, this means it's 10 cm from the wire carrying 1A current, on the side opposite to the wire carrying 2A current.

   Let's verify the other external region (to the right of Q, $x > 10$ cm).
   Let the distance from Q be $r_Q'$. The distance from P would be $d+r_Q' = 10+r_Q'$.

   $\frac{I_P}{10+r_Q'} = \frac{I_Q}{r_Q'}$

   $\frac{1}{10+r_Q'} = \frac{2}{r_Q'}$

   $r_Q' = 2(10+r_Q')$

   $r_Q' = 20 + 2r_Q'$

   $-r_Q' = 20$

   $r_Q' = -20$ cm. This is not a physically valid distance (distance must be positive), confirming no solution exists in this region.

The resultant magnetic field will be zero at a point 10 cm from wire P (carrying 1A current), on the side away from wire Q (carrying 2A current).

Explanation of the solution:
The magnetic field due to a straight current-carrying wire is $B = \frac{\mu_0 I}{2 \pi r}$. For the resultant magnetic field to be zero, the fields from individual wires must be equal in magnitude and opposite in direction.
When currents are in opposite directions, the zero-field point lies outside the wires, closer to the wire with the smaller current.
Let the distance from wire P (1A) be $r_P$ and from wire Q (2A) be $r_Q$.
Setting $B_P = B_Q$: $\frac{I_P}{r_P} = \frac{I_Q}{r_Q}$.
Since the point is outside and closer to P, $r_Q = d + r_P$, where $d=10$ cm is the separation.
$\frac{1}{r_P} = \frac{2}{10 + r_P}$
Solving for $r_P$: $10 + r_P = 2r_P \implies r_P = 10$ cm.
This means the point is 10 cm from wire P, on the side away from wire Q.