A 4 A current carrying loop consists of three identical quarter circles of radius 5 cm lying in the positive quadrants of the x-y, y-z and z-x planes with their centres at the origin joined together, value of B at the origin is

Options

A. `(mu_(0))/(10)(hati+hatj-hatk)T`
B. `(mu_(0))/(10)(-hati+hatj+hatk)T`
C. `(mu_(0))/(5)(hati+hatj+hatk)T`
D. `10mu_(0)(hati+hatj+hatk)T`

Explanation

To find the magnetic field \( B \) at the origin due to a current-carrying loop consisting of three identical quarter circles, we can follow these steps:

### Step 1: Understand the Configuration
The loop consists of three quarter circles located in the positive quadrants of the x-y, y-z, and z-x planes. Each quarter circle has a radius of 5 cm (0.05 m), and the current flowing through the loop is 4 A.

### Step 2: Determine the Magnetic Field Contribution from Each Quarter Circle
The magnetic field \( B \) at the center of a circular loop due to a current \( I \) is given by the formula:
\[ B = \frac{\mu_0 I}{4\pi r} \theta \]
where:
- \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \, \text{T m/A} \)),
- \( r \) is the radius of the loop,
- \( \theta \) is the angle subtended at the center (for a quarter circle, \( \theta = \frac{\pi}{2} \)).

### Step 3: Calculate the Magnetic Field for Each Quarter Circle
For each quarter circle, the magnetic field at the origin can be calculated as follows:

1. **For the quarter circle in the x-y plane**:
\[ B_{xy} = \frac{\mu_0 I}{4\pi r} \cdot \frac{\pi}{2} \hat{k} \]
Substituting \( I = 4 \, \text{A} \) and \( r = 0.05 \, \text{m} \):
\[ B_{xy} = \frac{(4\pi \times 10^{-7}) \cdot 4}{4\pi \cdot 0.05} \cdot \frac{\pi}{2} \hat{k} \]

2. **For the quarter circle in the y-z plane**:
\[ B_{yz} = \frac{\mu_0 I}{4\pi r} \cdot \frac{\pi}{2} \hat{i} \]

3. **For the quarter circle in the z-x plane**:
\[ B_{zx} = \frac{\mu_0 I}{4\pi r} \cdot \frac{\pi}{2} \hat{j} \]

### Step 4: Combine the Magnetic Fields
Now, we can combine the contributions from all three quarter circles:
\[ B = B_{xy} + B_{yz} + B_{zx} \]
Substituting the values we calculated:
\[ B = \left( \frac{\mu_0 I}{4\pi r} \cdot \frac{\pi}{2} \hat{k} + \frac{\mu_0 I}{4\pi r} \cdot \frac{\pi}{2} \hat{i} + \frac{\mu_0 I}{4\pi r} \cdot \frac{\pi}{2} \hat{j} \right) \]
Factoring out the common term:
\[ B = \frac{\mu_0 I}{4\pi r} \cdot \frac{\pi}{2} \left( \hat{i} + \hat{j} + \hat{k} \right) \]

### Step 5: Substitute the Values
Now substituting \( I = 4 \, \text{A} \), \( r = 0.05 \, \text{m} \):
\[ B = \frac{(4\pi \times 10^{-7}) \cdot 4}{4\pi \cdot 0.05} \cdot \frac{\pi}{2} \left( \hat{i} + \hat{j} + \hat{k} \right) \]
This simplifies to:
\[ B = \frac{(4 \times 10^{-7}) \cdot 4}{0.2} \cdot \frac{1}{2} \left( \hat{i} + \hat{j} + \hat{k} \right) \]
\[ B = (8 \times 10^{-6}) \left( \hat{i} + \hat{j} + \hat{k} \right) \text{ T} \]

### Final Result
Thus, the magnetic field at the origin is:
\[ B = 10 \mu_0 \left( \hat{i} + \hat{j} + \hat{k} \right) \text{ T} \]