A current of 200 $\mu A$ deflects the coil of a moving coil galvanometer through $60^{\circ}$. The current to cause deflection through $\frac{\pi}{10}$ radian is :
- 30 $\mu A$
- 120 $\mu A$
- 60 $\mu A$
- 180 $\mu A$
The Correct Option is C
Approach Solution - 1
To solve this problem, we need to understand the working principle of a moving coil galvanometer. The deflection in a moving coil galvanometer is directly proportional to the current flowing through it. This is based on the principle that the torque on a coil in a magnetic field is proportional to the current and the number of turns in the coil.
The deflection can be represented as:
\(\theta \propto I\)
Given:
- A current of 200 \(\mu A\) produces a deflection of \(60^{\circ}\).
- We need to find the current that causes a deflection through \(\frac{\pi}{10}\) radians.
First, convert \(60^{\circ}\) to radians:
\(60^{\circ} = \frac{60 \times \pi}{180} = \frac{\pi}{3} \text{ radians}\)
Using the proportionality, write the equation for both scenarios:
\(\frac{\pi}{3} \propto 200 \mu A\)
\(\frac{\pi}{10} \propto I_{\text{required}}\)
Equating the ratios as the proportionality constant is the same:
\(\frac{200}{I_{\text{required}}} = \frac{\frac{\pi}{3}}{\frac{\pi}{10}}\)
Simplify the right-hand side:
\(\frac{\pi}{3} \cdot \frac{10}{\pi} = \frac{10}{3}\)
Thus, the equation becomes:
\(\frac{200}{I_{\text{required}}} = \frac{10}{3}\)
Cross-multiply to solve for \(I_{\text{required}}\):
\(I_{\text{required}} = \frac{200 \times 3}{10} = 60 \mu A\)
Therefore, the current required to cause a deflection of \(\frac{\pi}{10}\) radians is 60 \(\mu A\).
Hence, the correct option is 60 \(\mu A\).
Approach Solution -2
Given:
- \( i_2 = 200 \, \mu A \),
- \( \theta_2 = 60^\circ = \frac{\pi}{3} \, \text{radians} \).
The deflection \( \theta \) is proportional to the current \( i \). Therefore:
\(\frac{i_1}{i_2} = \frac{\theta_1}{\theta_2}.\)
For \( \theta_1 = \frac{\pi}{10} \, \text{radians} \):
\(\frac{i_1}{200} = \frac{\frac{\pi}{10}}{\frac{\pi}{3}}.\)
Simplify:
\(\frac{i_1}{200} = \frac{3}{10} \implies i_1 = 200 \times \frac{3}{10} = 60 \, \mu A.\)
The Correct answer is: 60 $\mu A$
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