Question

Two long parallel wires carry currents \( I_1 \) and \( I_2 \) (\( I_1 > I_2 \)). When currents are flowing in the same direction, the magnetic field at a point midway between the wires is \( 6 \times 10^{-6} \, \text{T} \). If the direction of \( I_2 \) is reversed, the field at the midpoint becomes \( 3 \times 10^{-5} \, \text{T} \). The ratio \( I_1 : I_2 \) is

• A. 3 : 2
• B. 2 : 3
• C. 3 : 5
• D. 6 : 7

Hint:

When two currents flow in the same direction, their magnetic fields at a midpoint add up. When the currents flow in opposite directions, their magnetic fields subtract from each other.

Answer 1

The Correct Option is B

Step 1: Understanding the magnetic field due to current in long parallel wires.
The magnetic field at a point due to a long straight current-carrying wire is given by Ampère's law:
\[ B = \frac{\mu_0 I}{2 \pi r}, \]
where:
- \( B \) is the magnetic field at a point at a distance \( r \) from the wire,
- \( I \) is the current in the wire,
- \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \, \text{T m/A} \)),
- \( r \) is the perpendicular distance from the wire to the point where the field is being calculated.
For two parallel wires carrying currents \( I_1 \) and \( I_2 \), the total magnetic field at a point midway between the wires is the vector sum of the fields due to each wire.

Step 2: Magnetic field at the midpoint for currents in the same direction.
When the currents are flowing in the same direction, the magnetic fields due to both wires at the midpoint add up. Let the distance between the wires be \( d \). The magnetic field at the midpoint from each wire is:
\[ B_1 = \frac{\mu_0 I_1}{2 \pi \frac{d}{2}}, \quad B_2 = \frac{\mu_0 I_2}{2 \pi \frac{d}{2}}. \]
The total magnetic field at the midpoint is:
\[ B_{\text{total}} = B_1 + B_2 = \frac{\mu_0}{2 \pi \frac{d}{2}} (I_1 + I_2). \]
Given that this total field is \( 6 \times 10^{-6} \, \text{T} \), we have:
\[ \frac{\mu_0}{2 \pi \frac{d}{2}} (I_1 + I_2) = 6 \times 10^{-6}. \]

Step 3: Magnetic field at the midpoint for currents in opposite directions.
When the current in \( I_2 \) is reversed, the magnetic field due to \( I_2 \) at the midpoint opposes that due to \( I_1 \). Therefore, the total field becomes:
\[ B_{\text{total}} = B_1 - B_2 = \frac{\mu_0}{2 \pi \frac{d}{2}} (I_1 - I_2). \]
We are given that this total field is \( 3 \times 10^{-5} \, \text{T} \), so:
\[ \frac{\mu_0}{2 \pi \frac{d}{2}} (I_1 - I_2) = 3 \times 10^{-5}. \]

Step 4: Solving the system of equations.
We now have two equations:
1. \( \frac{\mu_0}{2 \pi \frac{d}{2}} (I_1 + I_2) = 6 \times 10^{-6} \),
2. \( \frac{\mu_0}{2 \pi \frac{d}{2}} (I_1 - I_2) = 3 \times 10^{-5} \).
Let \( k = \frac{\mu_0}{2 \pi \frac{d}{2}} \). The equations become:
1. \( k (I_1 + I_2) = 6 \times 10^{-6} \),
2. \( k (I_1 - I_2) = 3 \times 10^{-5} \).
Solving these two equations simultaneously, we first divide equation (2) by equation (1):
\[ \frac{k (I_1 - I_2)}{k (I_1 + I_2)} = \frac{3 \times 10^{-5}}{6 \times 10^{-6}}, \] \[ \frac{I_1 - I_2}{I_1 + I_2} = 5. \] This simplifies to: \[ I_1 - I_2 = 5 (I_1 + I_2), \] \[ I_1 - I_2 = 5 I_1 + 5 I_2. \]
Rearranging:
\[ I_1 - 5 I_1 = 5 I_2 + I_2, \] \[ -4 I_1 = 6 I_2, \] \[ \frac{I_1}{I_2} = \frac{6}{4} = 3:2. \]
Final Answer:
The ratio \( I_1 : I_2 \) is:
\[ \boxed{3:2}. \]