Question

Two tangent galvanometers A and B have coils of radii 8 cm and 16 cm respectively and resistance $8\Omega$ each. They are connected in parallel with a cell of emf 4 V and negligible internal resistance. The deflections produced in the tangent galvanometer A and B are 30° and 60° respectively. If A has 2 turns, then B must have:
A. 18 turns
B. 12 turns
C. 6 turns
D. 2 turns

Answer 1

To solve this problem, use the formula for current in tangent galvanometer. Use this same formula to find the current in tangent galvanometer A and galvanometer B. As these galvanometers are connected in parallel, same current will flow through both the galvanometers. So, compare the equation and obtain an expression. Substitute the given values in this obtained expression and find the number of turns in tangent galvanometer B.

Formula used:
$I=\dfrac { 2rH }{ { \mu }_{ 0 }N } \tan { \theta }$

Complete step-by-step solution:
Current in tangent galvanometer is given by,
$I=\dfrac { 2rH }{ { \mu }_{ 0 }N } \tan { \theta }$
Where,
r is the radius of the coil
H is the magnetic induction
N is the number of turns of the coil
$\theta$ is the angle of deflection
${\mu}_{0}$ is the permittivity of free space

So, the current in tangent galvanometer A is given by,
${I}_{A}=\dfrac { 2{r}_{A}H }{ { \mu }_{ 0 }{N}_{A} } \tan { {\theta}_{A} }$ …(1)
Similarly, current in tangent galvanometer B is given by,
${I}_{B}=\dfrac { 2{r}_{B}H }{ { \mu }_{ 0 }{N}_{B} } \tan { {\theta}_{B} }$ …(2)
Given,
${r}_{A}= 8 cm$
${r}_{B}= 16 cm$
${R}_{1}= 8 \Omega$
${R}_{2}= 8 \Omega$
${\theta}_{A}=30°$
${\theta}_{B}=60°$
${N}_{A}=2 \quad turns$
V= 4 V

It is given that both the resistors are connected in parallel. So, their equivalent resistance will be given by,
$\dfrac {1}{R}= \dfrac {1}{{R}_{1}}+ \dfrac {1}{{R}_{2}}$

Substituting values in above equation we get,
$\dfrac {1}{R}= \dfrac {1}{8}+ \dfrac {1}{8}$
$\Rightarrow \dfrac {1}{R}= \dfrac {2}{8}$
$\Rightarrow \dfrac {1}{R}= \dfrac {1}{4}$
$\Rightarrow R= 4 \Omega$
According to Ohm’s law,
$V=IR$

Substituting values in above equation we get,
$4=I \times 4$
$\Rightarrow I= 4 A$

This same current will flow through both the tangent galvanometers. Thus, equation. (1) will be equal to equation. (2)
$\therefore \dfrac { 2{r}_{A}H }{ { \mu }_{ 0 }{N}_{A} } \tan { {\theta}_{A} }=\dfrac { 2{r}_{B}H }{ { \mu }_{ 0 }{N}_{B} } \tan { {\theta}_{B} }$

Cancelling the common terms on both the sides we get,
$\dfrac { {r}_{A} }{ {N}_{A} } \tan { {\theta}_{A} }=\dfrac { {r}_{B} }{ {N}_{B} } \tan { {\theta}_{B} }$

Substituting values in above equation we get,
$\dfrac { 8\times \tan { { 30 }^{ ° } } }{ 2 } =\dfrac { 16\times \tan { { 60 }^{ ° } } }{ { N }_{ B } }$
$\Rightarrow \dfrac { 8\times 1 }{ 2\times \sqrt { 3 } } =\dfrac { 16\times \sqrt { 3 } }{ { N }_{ B } }$
$\Rightarrow {N}_{B}= \dfrac {16 \times \sqrt{3} \times 2 \sqrt {3}}{8}$
$\Rightarrow {N}_{B}= \dfrac {96}{8}$
$\Rightarrow {N}_{B}= 12 \quad turns$

Hence, the correct option is B.

Note: To solve this question, students must know about tangent galvanometer. If we want to increase the sensitivity of a tangent galvanometer, this can be done by increasing the number of turns of the coil or by decreasing the radius of the coil or the magnetic induction. Students must understand that if we increase the number of turns of the coil then the radius of the coil will not be the same for all the turns and the magnetic induction will not be uniform at the center.