The magnetic field at a point `x` on the axis of a small bar magnet is equal to the field at a point `y` on the equator of the same magnet. The ratio of the distances of `x` and `y` from the centre of the magnet is

Options

A. `2^(-3)`
B. `2^(-1//3)`
C. `2^(3)`
D. `2^(1//3)`

Explanation

To solve the problem, we need to find the ratio of the distances of points `x` and `y` from the center of a small bar magnet, given that the magnetic field at point `x` on the axis of the magnet is equal to the magnetic field at point `y` on the equator of the same magnet.

### Step-by-Step Solution:

1. **Understanding the Magnetic Field of a Bar Magnet**:
- The magnetic field \( B \) at a point on the axis of a bar magnet at a distance \( d \) from its center is given by:
\[ B_x = \frac{\mu_0}{4\pi} \cdot \frac{2M}{d^3} \]
where \( M \) is the magnetic moment of the magnet.

2. **Magnetic Field at Point `y` on the Equator**:
- The magnetic field \( B \) at a point on the equator of the magnet at a distance \( d \) from the center is given by:
\[ B_y = \frac{\mu_0}{4\pi} \cdot \frac{M}{2d^3} \]

3. **Setting the Fields Equal**:
- According to the problem, the magnetic fields at points `x` and `y` are equal:
\[ B_x = B_y \]
- Substituting the expressions for \( B_x \) and \( B_y \):
\[ \frac{\mu_0}{4\pi} \cdot \frac{2M}{d_x^3} = \frac{\mu_0}{4\pi} \cdot \frac{M}{2d_y^3} \]
- We can cancel \( \frac{\mu_0}{4\pi} \) from both sides:
\[ \frac{2M}{d_x^3} = \frac{M}{2d_y^3} \]

4. **Simplifying the Equation**:
- Dividing both sides by \( M \) (assuming \( M \neq 0 \)):
\[ \frac{2}{d_x^3} = \frac{1}{2d_y^3} \]
- Cross-multiplying gives:
\[ 2 \cdot 2d_y^3 = d_x^3 \]
\[ 4d_y^3 = d_x^3 \]

5. **Finding the Ratio of Distances**:
- Taking the cube root of both sides:
\[ d_x = 4^{1/3} d_y \]
- Therefore, the ratio of the distances \( \frac{d_x}{d_y} \) is:
\[ \frac{d_x}{d_y} = 4^{1/3} \]

### Final Answer:
The ratio of the distances of `x` and `y` from the center of the magnet is \( 4^{1/3} \).

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