A point charge `q=2muC` is at the origin. It has velocity `2 hati m//s`.Find the magnetic field at the following point in vector from (at the moment when the charged particle passes through the origin) :
(i) `(2,0,0)` , (ii) `(0,2,0)` , (iii) `(0,0,2)` , (iv) `(2,1,2)`
(v)Is the magnitude of the magnetic field on the circumference of the circle (in `yz` plane) `y^(2)+z^(2)=c^(2)` where `c^(2)` is a contant is same every where.Is it same in direction also.
(vI) Answer the above (vii) for the circle of same equation but in a plane `x=a` where `a` is a contant.

Answer

To solve the problem step by step, we will use the formula for the magnetic field \( \mathbf{B} \) created by a moving point charge. The formula is given by:

\[ \mathbf{B} = \frac{\mu_0}{4\pi} \frac{q \mathbf{v} \times \mathbf{r}}{r^3} \]

where:
- \( \mu_0 \) is the permeability of free space,
- \( q \) is the charge,
- \( \mathbf{v} \) is the velocity vector of the charge,
- \( \mathbf{r} \) is the position vector from the charge to the point where the magnetic field is being calculated,
- \( r \) is the magnitude of the position vector \( \mathbf{r} \).

Given:
- \( q = 2 \, \mu C = 2 \times 10^{-6} \, C \)
- \( \mathbf{v} = 2 \hat{i} \, m/s \)

Now, we will calculate the magnetic field at each specified point.

### (i) At point (2, 0, 0)

1. **Position Vector**: \( \mathbf{r} = 2 \hat{i} \)
2. **Magnitude**: \( r = |\mathbf{r}| = 2 \)
3. **Cross Product**: \( \mathbf{v} \times \mathbf{r} = (2 \hat{i}) \times (2 \hat{i}) = 0 \)
4. **Magnetic Field**:
\[ \mathbf{B} = \frac{\mu_0}{4\pi} \frac{q \cdot 0}{2^3} = 0 \]

### (ii) At point (0, 2, 0)

1. **Position Vector**: \( \mathbf{r} = 2 \hat{j} \)
2. **Magnitude**: \( r = |\mathbf{r}| = 2 \)
3. **Cross Product**:
\[ \mathbf{v} \times \mathbf{r} = (2 \hat{i}) \times (2 \hat{j}) = 4 \hat{k} \]
4. **Magnetic Field**:
\[ \mathbf{B} = \frac{\mu_0}{4\pi} \frac{(2 \times 10^{-6}) \cdot (4 \hat{k})}{2^3} = \frac{\mu_0 \cdot 8 \hat{k}}{32 \pi} = \frac{\mu_0}{4\pi} \cdot \frac{1}{4} \hat{k} \]

### (iii) At point (0, 0, 2)

1. **Position Vector**: \( \mathbf{r} = 2 \hat{k} \)
2. **Magnitude**: \( r = |\mathbf{r}| = 2 \)
3. **Cross Product**:
\[ \mathbf{v} \times \mathbf{r} = (2 \hat{i}) \times (2 \hat{k}) = -4 \hat{j} \]
4. **Magnetic Field**:
\[ \mathbf{B} = \frac{\mu_0}{4\pi} \frac{(2 \times 10^{-6}) \cdot (-4 \hat{j})}{2^3} = -\frac{\mu_0 \cdot 8 \hat{j}}{32 \pi} = -\frac{\mu_0}{4\pi} \cdot \frac{1}{4} \hat{j} \]

### (iv) At point (2, 1, 2)

1. **Position Vector**: \( \mathbf{r} = 2 \hat{i} + 1 \hat{j} + 2 \hat{k} \)
2. **Magnitude**:
\[ r = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4 + 1 + 4} = 3 \]
3. **Cross Product**:
\[ \mathbf{v} \times \mathbf{r} = (2 \hat{i}) \times (2 \hat{i} + 1 \hat{j} + 2 \hat{k}) = 2 \hat{i} \times 1 \hat{j} + 2 \hat{i} \times 2 \hat{k} = 2 \hat{k} - 4 \hat{j} \]
4. **Magnetic Field**:
\[ \mathbf{B} = \frac{\mu_0}{4\pi} \frac{(2 \times 10^{-6}) (2 \hat{k} - 4 \hat{j})}{3^3} = \frac{\mu_0}{4\pi} \frac{(4 \hat{k} - 8 \hat{j}) \times 10^{-6}}{27} \]

### (v) For the circle in the \( yz \) plane \( y^2 + z^2 = c^2 \)

The magnetic field will have the same magnitude at every point on the circumference since the distance \( r \) remains constant. The direction will vary depending on the angle \( \theta \) in the \( yz \) plane.

### (vi) For the circle in the plane \( x = a \)

The analysis is similar to part (v). The magnetic field will have the same magnitude at every point on the circumference, and the direction will also vary depending on the position on the circle.