# Determine the Ratio of (A) Current Sensitivity and (B) Voltage Sensitivity of M2 And M1. - Physics

Numerical

Two moving coil meters, M1 and M2 have the following particulars:

R1 = 10 Ω, N1 = 30,

A1 = 3.6 × 10–3 m2, B1 = 0.25 T

R2 = 14 Ω, N2 = 42,

A2 = 1.8 × 10–3 m2, B2 = 0.50 T

(The spring constants are identical for the two meters).

Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M2 and M1.

#### Solution

For moving coil meter M1:

Resistance, R1 = 10 Ω

Number of turns, N1 = 30

Area of cross-section, A1 = 3.6 × 10–3 m2

Magnetic field strength, B1 = 0.25 T

Spring constant K1 = K

For moving coil meter M2:

Resistance, R2 = 14 Ω

Number of turns, N2 = 42

Area of cross-section, A2 = 1.8 × 10–3 m2

Magnetic field strength, B2 = 0.50 T

Spring constant, K2 = K

(a) Current sensitivity of M1 is given as:

"I"_("s"_1) = ("N"_1"B"_1"A"_1)/"K"_1

And, current sensitivity of M2 is given as:

"I"_("s"_2) = ("N"_2"B"_2"A"_2)/"K"_2

∴ Ratio "I"_("s"_2)/"I"_("s"_1) = ("N"_2"B"_2"A"_2"K"_1)/("K"_2"N"_1"B"_1"A"_1)

= (42 xx 0.5 xx 1.8 xx 10^-3 xx "K")/("K" xx 30 xx 0.25 xx 3.6 xx 10^-3) = 1.4

Hence, the ratio of current sensitivity of M2 to M1 is 1.4.

(b) Voltage sensitivity for M2 is given as:

"V"_("s"_2) = ("N"_2"B"_2"A"_2)/("K"_2"R"_2)

And, voltage sensitivity for M1 is given as:

"V"_("s"_1) = ("N"_1"B"_1"A"_1)/("K"_1"R"_1)

∴ Ratio "V"_("s"_2)/"V"_("s"_1) = ("N"_2"B"_2"A"_2"K"_1"R"_1)/("K"_2"R"_2"N"_1"B"_1"A"_1)

= (42 xx 0.5 xx 1.8 xx 10^-3 xx 10 xx"K")/("K" xx 14 xx 30 xx 0.25 xx 3.6 xx 10^-3) = 1

Hence, the ratio of voltage sensitivity of M2 to M1 is 1.

Concept: Moving Coil Galvanometer
Is there an error in this question or solution?
Chapter 4: Moving Charges and Magnetism - Exercise [Page 169]

#### APPEARS IN

NCERT Physics Part 1 and 2 Class 12
Chapter 4 Moving Charges and Magnetism
Exercise | Q 4.10 | Page 169
NCERT Class 12 Physics Textbook
Chapter 4 Moving Charges and Magnetism
Exercise 2 | Q 10 | Page 169

Share