Q.Two moving coil meters, $M_1$ and $M_2$ have the following particulars:
$R_1 = 10\ \Omega$, $N_1 = 30$, $A_1 = 3.6 \times 10^{-3}\ \text{m}^2$, $B_1 = 0.25\ \text{T}$
$R_2 = 14\ \Omega$, $N_2 = 42$, $A_2 = 1.8 \times 10^{-3}\ \text{m}^2$, $B_2 = 0.50\ \text{T}$
(The spring constants are identical for the two meters). Determine the ratio of

Concept: Ammeter Loading Effect — but here we directly compare sensitivities using construction parameters.

Reasoning:

1. Current sensitivity of a moving coil meter is $S_I = \frac{NBA}{k}$, where $k$ is the spring constant (same for both). So the ratio depends only on $NBA$.

   For $M_1$: $N_1 B_1 A_1 = 30 \times 0.25 \times 3.6 \times 10^{-3} = 27 \times 10^{-3}$

   For $M_2$: $N_2 B_2 A_2 = 42 \times 0.50 \times 1.8 \times 10^{-3} = 37.8 \times 10^{-3}$

   Ratio $\frac{S_{I2}}{S_{I1}} = \frac{37.8}{27} = 1.4$ …

A moving coil meter works by passing current through a coil in a magnetic field. The torque produced is $\tau = N B I A$, where $N$ is the number of turns, $B$ the magnetic field, $I$ the current, and $A$ the area of the coil. This torque is opposed by the spring, which gives a restoring torque $\tau_s = k \theta$, with $k$ the spring constant (same for both meters here). At equilibrium, $N B I A = k \theta$, so the deflection $\theta$ is proportional to $I$.

Current sensitivity is defined as deflection per unit current: $S_I = \theta / I = N B A / k$. Since $k$ is identical for $M_1$ and $M_2$, the ratio of current sensitivities is simply the ratio of $NBA$ products.

Voltage sensitivity is deflection per unit voltage. If the meter has resistance $R$, then $I = V/R$, so $\theta = (N B A / k) \cdot (V/R)$. Hence $S_V = \theta / V = N B A / (k R)$. Again, $k$ cancels in the ratio.

Let’s compute step by step.

1. Current sensitivity ratio

   For $M_1$: $N_1 B_1 A_1 = 30 \times 0.25 \times (3.6 \times 10^{-3})$

   $= 30 \times 0.25 = 7.5$, then $7.5 \times 3.6 \times 10^{-3} = 27 \times 10^{-3} = 0.027$

   For $M_2$: $N_2 B_2 A_2 = 42 \times 0.50 \times (1.8 \times 10^{-3})$

   $= 42 \times 0.50 = 21$, then $21 \times 1.8 \times 10^{-3} = 37.8 \times 10^{-3} = 0.0378$

   Ratio $\frac{S_{I2}}{S_{I1}} = \frac{0.0378}{0.027} = 1.4$

2. Voltage sensitivity ratio

   For $M_1$: $\frac{N_1 B_1 A_1}{R_1} = \frac{0.027}{10} = 0.0027$ …

Method: Direct Formula Substitution Method

This method uses the standard formulas for current sensitivity and voltage sensitivity of a moving coil galvanometer, then directly substitutes the given data.

Step 1: Recall the formulas

• Current sensitivity ($S_I$) = $\frac{NBA}{k}$

  (deflection per unit current)

• Voltage sensitivity ($S_V$) = $\frac{NBA}{kR}$

  (deflection per unit voltage)

Here, $k$ is the spring constant (same for both meters).

Step 2: Write the ratio for current sensitivity

We need $\frac{S_{I2}}{S_{I1}}$

Substitute values:

Simplify step-by-step:

• Numerator: $42 \times 0.50 = 21$, then $21 \times 1.8 \times 10^{-3} = 37.8 \times 10^{-3}$
• Denominator: $30 \times 0.25 = 7.5$, then $7.5 \times 3.6 \times 10^{-3} = 27 \times 10^{-3}$

Answer (a): $\boxed{1.4}$

Step 3: Write the ratio for voltage sensitivity …

Common Mistakes: Ammeter Loading Effect & Sensitivity Problems

Students often confuse current sensitivity and voltage sensitivity — and then make avoidable errors in ratio problems. Here are the most frequent mistakes and how to avoid each.

✗ Mistake 1: Confusing the formulas for current sensitivity and voltage sensitivity

What students do wrong:

They swap the formulas or use the same formula for both.

Correct understanding:

• Current sensitivity ($S_I$) = deflection per unit current

where $k$ is the spring constant (same for both meters here).

• Voltage sensitivity ($S_V$) = deflection per unit voltage

How to avoid:

Write both formulas side-by-side before starting. Remember: voltage sensitivity = current sensitivity divided by resistance.

✗ Mistake 2: Forgetting that spring constants are identical

What students do wrong:

They try to find $k$ numerically or cancel it incorrectly.

Correct approach:

Since $k$ is the same for $M_1$ and $M_2$, it cancels out in the ratio. You only need $N$, $B$, $A$, and $R$.

How to avoid:

Always check which quantities are common between the meters before writing ratios.

✗ Mistake 3: Incorrect ratio direction

What students do wrong:

They compute $\frac{S_{M1}}{S_{M2}}$ instead of $\frac{S_{M2}}{S_{M1}}$.

Correct approach:

The question asks: ratio of $M_2$ and $M_1$ → $\frac{S_{M2}}{S_{M1}}$.

How to avoid:

Read carefully: “ratio of A and B” means $\frac{A}{B}$.

✗ Mistake 4: Arithmetic errors in multiplication/division

What students do wrong:

They mess up the product $N \times B \times A$ or forget to square units.

Correct calculation for part (a):

Substitute:

Simplify step-by-step:

• Numerator: $42 \times 0.50 = 21$, then $21 \times 1.8 = 37.8$, then $\times 10^{-3}$
• Denominator: $30 \times 0.25 = 7.5$, then $7.5 \times 3.6 = 27$, then $\times 10^{-3}$

How to avoid:

Do one multiplication at a time. Cancel $10^{-3}$ early.

✗ Mistake 5: Forgetting to include resistance in voltage sensitivity ratio …