The relation between voltage sensitivity (σᵥ) and current sensitivity (σᵢ) of a moving coil galvanometer is (resistance of galvanometer is G):
(A) (σᵢ / G) = σᵥ
(B) (σᵥ / G) = σᵢ
(C) (G / σᵥ) = σᵢ
(D) None of the above

To solve the given problem, we need to determine the relationship between the voltage sensitivity (σv) and current sensitivity (σi) of a moving coil galvanometer. Here's the detailed explanation:

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### Definitions:
1. **Voltage sensitivity (σv):** Voltage sensitivity is defined as the deflection per unit voltage applied to the galvanometer:
\[
\sigma_v = \frac{\theta}{V}
\]
where θ is the deflection and V is the applied voltage.

2. **Current sensitivity (σi):** Current sensitivity is defined as the deflection per unit current passing through the galvanometer:
\[
\sigma_i = \frac{\theta}{I}
\]
where I is the current through the galvanometer.

3. The resistance of the galvanometer is denoted as \( G \).

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### Voltage and Current Relationship:
The applied voltage \( V \) and current \( I \) through the galvanometer are related by Ohm's law:
\[
V = I \cdot G
\]
From this, we can express \( I \) as:
\[
I = \frac{V}{G}
\]

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### Relationship Between σv and σi:
Substitute \( I = \frac{V}{G} \) into the expression for current sensitivity \( \sigma_i = \frac{\theta}{I} \):
\[
\sigma_i = \frac{\theta}{I} = \frac{\theta}{\frac{V}{G}} = \frac{\theta \cdot G}{V}
\]
Now compare this with the expression for voltage sensitivity:
\[
\sigma_v = \frac{\theta}{V}
\]
From the above two equations, it is evident that:
\[
\sigma_i = \sigma_v \cdot G
\]
or equivalently:
\[
\frac{\sigma_v}{G} = \sigma_i
\]

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### Correct Option:
The correct relationship is:
\[
\frac{\sigma_v}{G} = \sigma_i
\]
Thus, the correct option is **B**.