Two short bar magnets have equal pole stengths but one is twice as long as the other one . The short magnet is placed at a distance of 30 cm in tan A position form the compass needle . The longer magnet must be placed on the other side of the compass needle for no deflection at a distance of .

Options

A. 30 cm
B. `30 xx (2)^(2//3) cm `
C. `30 xx (2)^(1//2) cm `
D. `30 xx 2 cm `

Explanation

To solve the problem, we need to find the distance \( d \) at which the longer magnet must be placed so that there is no deflection of the compass needle when one magnet is placed at a distance of 30 cm.

### Step-by-step Solution:

1. **Identify the Given Data:**
- Let the pole strength of both magnets be \( m \).
- Let the length of the shorter magnet be \( L \) and the longer magnet be \( 2L \).
- The distance of the shorter magnet from the compass needle is \( 30 \) cm.

2. **Understand the Magnetic Field Contributions:**
- The magnetic field \( B \) due to a bar magnet at a distance \( r \) on its axial line is given by:
\[ B_{\text{axial}} = \frac{\mu_0 m}{4 \pi r^3} \]
- The magnetic field \( B \) due to a bar magnet at a distance \( r \) on its end-on position is given by:
\[ B_{\text{end-on}} = \frac{\mu_0 (2m)}{4 \pi d^3} \]
- Here, \( d \) is the distance of the longer magnet from the compass needle.

3. **Set Up the Equation for No Deflection:**
- For no deflection of the compass needle, the magnetic field due to the shorter magnet must equal the magnetic field due to the longer magnet:
\[ B_{\text{axial}} = B_{\text{end-on}} \]
- Substituting the expressions for \( B \):
\[ \frac{\mu_0 m}{4 \pi (30)^3} = \frac{\mu_0 (2m)}{4 \pi d^3} \]

4. **Simplify the Equation:**
- Cancel out common terms:
\[ \frac{1}{30^3} = \frac{2}{d^3} \]
- Rearranging gives:
\[ d^3 = 2 \times 30^3 \]

5. **Calculate \( d \):**
- Taking the cube root:
\[ d = 30 \times 2^{1/3} \]

6. **Final Calculation:**
- The value of \( 2^{1/3} \) is approximately \( 1.26 \):
\[ d \approx 30 \times 1.26 \approx 37.8 \text{ cm} \]

### Conclusion:
The longer magnet must be placed at a distance of approximately \( 37.8 \) cm from the compass needle for no deflection.