At a certain place, a magnet makes 30 oscillations per minute. At another place, the magnetic field is double. What is the its time period ?

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript.

### Step-by-Step Solution:

1. **Understanding the Given Information:**
- The magnet makes 30 oscillations per minute at the initial place.
- We need to find the time period when the magnetic field is doubled.

2. **Calculate the Initial Time Period:**
- The number of oscillations per minute is given as 30.
- To find the time period (T_initial), we convert oscillations per minute to seconds:
\[ \text{Time Period} (T) = \frac{60 \text{ seconds}}{\text{Number of oscillations per minute}} = \frac{60}{30} = 2 \text{ seconds} \]
- Thus, \( T_{\text{initial}} = 2 \text{ seconds} \).

3. **Understanding the Relationship Between Time Period and Magnetic Field:**
- The time period (T) of a magnet is related to the magnetic field (B) by the formula:
\[ T \propto \frac{1}{\sqrt{B}} \]
- This means that if the magnetic field changes, the time period will also change according to this relationship.

4. **Setting Up the Equation:**
- Let \( B_{\text{initial}} = B \) and \( B_{\text{final}} = 2B \).
- According to the proportionality, we can write:
\[ T_{\text{initial}} \sqrt{B_{\text{initial}}} = T_{\text{final}} \sqrt{B_{\text{final}}} \]
- Substituting the values:
\[ 2 \sqrt{B} = T_{\text{final}} \sqrt{2B} \]

5. **Simplifying the Equation:**
- Dividing both sides by \( \sqrt{B} \):
\[ 2 = T_{\text{final}} \sqrt{2} \]

6. **Solving for the Final Time Period:**
- Rearranging the equation to find \( T_{\text{final}} \):
\[ T_{\text{final}} = \frac{2}{\sqrt{2}} = \sqrt{2} \text{ seconds} \]

7. **Calculating the Numerical Value:**
- The numerical value of \( \sqrt{2} \) is approximately 1.414 seconds.
- Therefore, \( T_{\text{final}} \approx 1.414 \text{ seconds} \).

### Final Answer:
The time period when the magnetic field is doubled is approximately **1.414 seconds**.

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