Question

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}},{B_1} = 0.25T \\
  {R_2} = 14\Omega ,{N_2} = 42 \\
  {A_2} = 1.8 \times {10^{ - 3}},{B_2} = 0.50T \\
 $
(The spring constants are identical for the two meters).
Determine the ratio of:
A. Current sensitivity and
B. Voltage Sensitivity of ${M_2}$ and ${M_1}$ .

Answer 1

Hint: In this question first we will find the current sensitivity of both the moving coil meters and then find the ratio similarly we will find out the voltage sensitivity of both the moving coil meters and hence we can find the solution.

Formula used:
We know that, current sensitivity in a moving coil galvanometer is given by,
$I = \dfrac{{BAN}}{K}$
And
We know that, voltage sensitivity in a moving coil galvanometer is given by,
$V = \dfrac{{BAN}}{{KR}}$
Where,
$I$ is current,
$B$ is the magnetic field,
$A$ is the area of cross section,
$N$ is the number of turns and
$R$ is the resistance.

Complete step by step answer:
For moving coil meter ${M_1}$ ,
Resistance is $10\Omega $ ,
Number of turns is $30$ ,
Area of cross section is $3.6 \times {10^{ - 3}}{m^2}$ ,
Magnetic field strength is $0.25T$ and
Spring constant is $K$
And
For moving coil meter ${M_2}$ ,
Resistance is $14\Omega $ ,
Number of turns is $42$ ,
Area of cross section is $1.8 \times {10^{ - 3}}{m^2}$ ,
Magnetic field strength is $0.50T$ and
Spring constant is $K$
As we know that the current sensitivity of a moving coil galvanometer is
$I = \dfrac{{BAN}}{K}$
Therefore, the ratio of their current sensitivity,
$\dfrac{{{I_{M1}}}}{{{I_{M2}}}} = \dfrac{{\dfrac{{{B_1}{A_1}{N_1}}}{{{K_1}}}}}{{\dfrac{{{B_2}{A_2}{N_2}}}{{{K_2}}}}}$
$
   \Rightarrow \dfrac{{{I_{M1}}}}{{{I_{M2}}}} = \dfrac{{42 \times 0.5 \times 1.8 \times {{10}^{ - 3}} \times K}}{{30 \times 0.25 \times 3.6 \times {{10}^{ - 3}} \times K}} \\
   \Rightarrow \dfrac{{{I_{M1}}}}{{{I_{M2}}}} = 1.4 \\
 $
Hence, we can say that the ratio of current sensitivity from ${M_2}$ to ${M_1}$ is $1.4$ .
Now, we will calculate the ratio of voltage sensitivity:
As we know that the voltage sensitivity of a moving coil galvanometer is
$V = \dfrac{{BAN}}{{KR}}$
Therefore, the ratio of their voltage sensitivity,
$\dfrac{{{V_{M1}}}}{{{V_{M2}}}} = \dfrac{{\dfrac{{{B_1}{A_1}{N_1}}}{{{R_1}{K_1}}}}}{{\dfrac{{{B_2}{A_2}{N_2}}}{{{R_1}{K_2}}}}}$
$
   \Rightarrow \dfrac{{{V_{M1}}}}{{{V_{M2}}}} = \dfrac{{42 \times 0.5 \times 1.8 \times {{10}^{ - 3}} \times 10 \times K}}{{14 \times 30 \times 0.25 \times 3.6 \times {{10}^{ - 3}} \times K}} \\
   \Rightarrow \dfrac{{{V_{M1}}}}{{{V_{M2}}}} = 1 \\
 $
Hence, we can say that the ratio of voltage sensitivity from ${M_2}$ to ${M_1}$ is $1$ .

Note: In this type of question, one must remember the formula of current sensitivity of a moving coil galvanometer and voltage sensitivity of a moving coil galvanometer to answer the formula, a galvanometer is a device that measures or detects small currents with appropriate modification. It can be converted into ammeter to measure the currents in the order of an ampere or millimetre or in the range of milliamperes or microammeter to measure microampere current. And note that the principle of moving coil galvanometer is a torque on a current loop placed in a magnetic field.