A wire carrying current `I` and other carrying `2I` in the sam direction produce a magnetic field `B` at the midpoint. What will be the field when `2I` wire is swiched off?

Options

A. `B//2`
B. `2B`
C. `B`
D. `4B`

Explanation

To solve the problem step by step, we will analyze the magnetic fields produced by the two wires at the midpoint and then determine the effect of switching off the current in the second wire.

### Step 1: Understand the Setup
We have two parallel wires:
- Wire 1 carries current \( I \).
- Wire 2 carries current \( 2I \).
Both currents flow in the same direction.

### Step 2: Magnetic Field due to Each Wire
The magnetic field \( B \) produced by a long straight wire carrying current \( I \) at a distance \( r \) from the wire is given by the formula:
\[ B = \frac{\mu_0 I}{2 \pi r} \]
where \( \mu_0 \) is the permeability of free space.

### Step 3: Calculate Magnetic Field at Midpoint
Let’s denote the distance from each wire to the midpoint as \( r \).
- The magnetic field due to wire 1 (current \( I \)) at the midpoint is:
\[ B_1 = \frac{\mu_0 I}{2 \pi r} \]
- The magnetic field due to wire 2 (current \( 2I \)) at the midpoint is:
\[ B_2 = \frac{\mu_0 (2I)}{2 \pi r} = \frac{\mu_0 I}{\pi r} \]

### Step 4: Determine the Net Magnetic Field
Since both currents are in the same direction, the magnetic fields produced by each wire at the midpoint will add up:
\[ B_{\text{net}} = B_1 + B_2 = \frac{\mu_0 I}{2 \pi r} + \frac{\mu_0 I}{\pi r} \]
To combine these, we can express \( B_2 \) with a common denominator:
\[ B_{\text{net}} = \frac{\mu_0 I}{2 \pi r} + \frac{2\mu_0 I}{2 \pi r} = \frac{3\mu_0 I}{2 \pi r} \]

### Step 5: Effect of Switching Off the Second Wire
When we switch off the current in the second wire (current \( 2I \)), only the first wire (current \( I \)) will contribute to the magnetic field at the midpoint. Thus, the magnetic field will be:
\[ B' = B_1 = \frac{\mu_0 I}{2 \pi r} \]

### Step 6: Relate \( B' \) to \( B \)
From our earlier calculations, we found that:
\[ B = \frac{3\mu_0 I}{2 \pi r} \]
Now, we can express \( B' \) in terms of \( B \):
\[ B' = \frac{\mu_0 I}{2 \pi r} = \frac{1}{3} B \]

### Conclusion
When the wire carrying current \( 2I \) is switched off, the magnetic field at the midpoint will be one-third of the original magnetic field \( B \) produced by both wires.

### Final Answer
The magnetic field when the wire carrying \( 2I \) is switched off is \( \frac{1}{3} B \).

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