A current of 1A is flowing along positive x-axis through a straight wire of length 0.5m placed in a region of a magnetic field given by `vec(B)= (2 hat(i) + 4hat(j))`T. The magnitude and the direction of the force experienced by the wire respectively are

Options

A. `sqrt18N`, along positive z-axis
B. `sqrt20N`, along positive x-axis
C. 2N , along positive z-axis
D. 4N, along positive y-axis

Explanation

To find the magnitude and direction of the force experienced by a straight wire carrying a current in a magnetic field, we can use the formula:

\[ \vec{F} = I \vec{L} \times \vec{B} \]

Where:
- \(\vec{F}\) is the force on the wire,
- \(I\) is the current,
- \(\vec{L}\) is the length vector of the wire,
- \(\vec{B}\) is the magnetic field vector.

### Step 1: Identify the given values
- Current, \(I = 1 \, \text{A}\)
- Length of the wire, \(L = 0.5 \, \text{m}\) (along the positive x-axis)
- Magnetic field, \(\vec{B} = 2 \hat{i} + 4 \hat{j} \, \text{T}\)

### Step 2: Write the length vector
Since the wire is along the positive x-axis, we can express the length vector as:
\[ \vec{L} = 0.5 \hat{i} \, \text{m} \]

### Step 3: Calculate the cross product \(\vec{L} \times \vec{B}\)
Using the formula for the cross product:
\[ \vec{F} = I (\vec{L} \times \vec{B}) = 1 \times (0.5 \hat{i} \times (2 \hat{i} + 4 \hat{j})) \]

### Step 4: Expand the cross product
Using the distributive property:
\[ \vec{L} \times \vec{B} = 0.5 \hat{i} \times (2 \hat{i}) + 0.5 \hat{i} \times (4 \hat{j}) \]
Calculating each term:
- \(\hat{i} \times \hat{i} = 0\)
- \(\hat{i} \times \hat{j} = \hat{k}\)

So we have:
\[ \vec{L} \times \vec{B} = 0 + 0.5 \times 4 \hat{k} = 2 \hat{k} \]

### Step 5: Calculate the force
Now substituting back into the force equation:
\[ \vec{F} = I (\vec{L} \times \vec{B}) = 1 \times (2 \hat{k}) = 2 \hat{k} \, \text{N} \]

### Step 6: Determine the magnitude and direction
- The magnitude of the force is \(2 \, \text{N}\).
- The direction of the force is along the positive z-axis (since \(\hat{k}\) indicates the z-direction).

### Final Answer
The magnitude and direction of the force experienced by the wire are:
- Magnitude: \(2 \, \text{N}\)
- Direction: Along the positive z-axis.

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