Same current i is flowing in the three infinitely long wires along positive x-,y- and z-directions. The magnetic filed at a point (0,0,-a) would be

Options

A. `(mu_(0)i)/(2pia)(hatj-hati)`
B. `(mu_(0)i)/(2pia)(hati+hatj)`
C. `(mu_(0)i)/(2pia)(hati-hatj)`
D. `(mu_(0)i)/(2pia)(hati-hatj+hatk)`

Answer

To find the magnetic field at the point (0, 0, -a) due to three infinitely long wires carrying the same current \( i \) along the positive x-, y-, and z-directions, we can follow these steps:

### Step 1: Identify the Magnetic Field Contributions
The magnetic field at a point due to a long straight wire carrying current can be calculated using the right-hand rule and the formula:
\[ B = \frac{\mu_0 i}{2 \pi r} \]
where \( r \) is the perpendicular distance from the wire to the point where the magnetic field is being calculated.

### Step 2: Analyze Each Wire
1. **Wire along the x-direction**: The wire is at the origin and extends along the x-axis. The point (0, 0, -a) is at a distance \( r = a \) from this wire. Using the right-hand rule, the magnetic field \( B_x \) at point (0, 0, -a) will be directed in the positive y-direction (j-cap).
\[ B_x = \frac{\mu_0 i}{2 \pi a} \hat{j} \]

2. **Wire along the y-direction**: Similarly, the wire along the y-axis is also at the origin. The point (0, 0, -a) is at a distance \( r = a \) from this wire. Using the right-hand rule, the magnetic field \( B_y \) at point (0, 0, -a) will be directed in the negative x-direction (-i-cap).
\[ B_y = -\frac{\mu_0 i}{2 \pi a} \hat{i} \]

3. **Wire along the z-direction**: The magnetic field at the point (0, 0, -a) due to the wire along the z-direction is zero because the point lies directly on the wire itself.
\[ B_z = 0 \]

### Step 3: Calculate the Net Magnetic Field
The net magnetic field \( B_{net} \) at the point (0, 0, -a) is the vector sum of the contributions from the three wires:
\[ B_{net} = B_x + B_y + B_z \]
Substituting the values we found:
\[ B_{net} = \frac{\mu_0 i}{2 \pi a} \hat{j} - \frac{\mu_0 i}{2 \pi a} \hat{i} + 0 \]
Thus,
\[ B_{net} = \frac{\mu_0 i}{2 \pi a} \hat{j} - \frac{\mu_0 i}{2 \pi a} \hat{i} \]

### Final Answer
The magnetic field at the point (0, 0, -a) is:
\[ B_{net} = \frac{\mu_0 i}{2 \pi a} \hat{j} - \frac{\mu_0 i}{2 \pi a} \hat{i} \]