Two short magnet having magnetic moment in the ratio `27:8`, when placed on the opposite sides of a deflection magnetometer, produce no deflection. If the distance of the weaker magnet is `0.12m` from the centre of deflection magnetometer, the distance of the stronger magnet from the centre is

Options

A. `0.06m`
B. `0.08m`
C. `0.12m`
D. `0.18m`

Answer

To solve the problem, we need to find the distance of the stronger magnet from the center of the deflection magnetometer, given the magnetic moments of the two magnets and the distance of the weaker magnet.

### Step-by-step Solution:

1. **Define the Magnetic Moments**:
Let the magnetic moment of the stronger magnet be \( M_1 \) and that of the weaker magnet be \( M_2 \). According to the problem, the ratio of their magnetic moments is given as:
\[ \frac{M_1}{M_2} = \frac{27}{8} \]

2. **Assign the Distance**:
The distance of the weaker magnet from the center of the magnetometer is given as \( d_2 = 0.12 \, \text{m} \). Let the distance of the stronger magnet from the center be \( d_1 \) (which we need to find).

3. **Magnetic Field Calculation**:
The magnetic field \( B \) produced by a magnetic dipole at a distance \( d \) is given by:
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{2M}{d^3} \]
Therefore, the magnetic field due to the stronger magnet at distance \( d_1 \) is:
\[ B_1 = \frac{\mu_0}{4\pi} \cdot \frac{2M_1}{d_1^3} \]
And for the weaker magnet at distance \( d_2 \):
\[ B_2 = \frac{\mu_0}{4\pi} \cdot \frac{2M_2}{d_2^3} \]

4. **Set the Magnetic Fields Equal**:
Since there is no deflection, the magnetic fields produced by both magnets must be equal:
\[ B_1 = B_2 \]
This leads to:
\[ \frac{2M_1}{d_1^3} = \frac{2M_2}{d_2^3} \]
The \( \frac{\mu_0}{4\pi} \) and 2 cancel out, simplifying to:
\[ \frac{M_1}{d_1^3} = \frac{M_2}{d_2^3} \]

5. **Substituting the Ratio of Magnetic Moments**:
Substitute \( M_1 = \frac{27}{8} M_2 \) into the equation:
\[ \frac{\frac{27}{8} M_2}{d_1^3} = \frac{M_2}{(0.12)^3} \]
Cancel \( M_2 \) from both sides:
\[ \frac{27}{8 d_1^3} = \frac{1}{(0.12)^3} \]

6. **Cross-Multiply and Solve for \( d_1^3 \)**:
Cross-multiplying gives:
\[ 27 \cdot (0.12)^3 = 8 \cdot d_1^3 \]
Calculate \( (0.12)^3 \):
\[ (0.12)^3 = 0.001728 \]
Thus:
\[ 27 \cdot 0.001728 = 8 \cdot d_1^3 \]
\[ 0.046656 = 8 d_1^3 \]
\[ d_1^3 = \frac{0.046656}{8} = 0.005832 \]

7. **Taking the Cube Root**:
Now, take the cube root to find \( d_1 \):
\[ d_1 = \sqrt[3]{0.005832} \approx 0.181 \]
Rounding gives:
\[ d_1 \approx 0.18 \, \text{m} \]

### Final Answer:
The distance of the stronger magnet from the center of the deflection magnetometer is approximately \( 0.18 \, \text{m} \).

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