Question

A Helmholtz galvanometer has coils of radius $\dfrac{11}{\sqrt{5}}cm$ and the number of turns $70$. Calculate the current through the coil which produces a deflection of $45{}^\circ$. Given magnetic intensity due to earth along horizontal is $H=\dfrac{80}{\pi }A{{m}^{-1}}$.
$\begin{aligned} & A)1.9\times {{10}^{7}}mA \\\ & B)10mA \\\ & C)15mA \\\ & D)25mA \\\ \end{aligned}$

Answer 1

Helmholtz galvanometer is a device which is used to produce magnetic fields with the help of electric current. It consists of two current carrying coils, kept on the same axis. Magnetic field produced is proportional to the deflection in the galvanometer. The current flowing through these coils can easily be determined if we know the values of the magnetic field generated between the coils, equal number of turns in these coils and equal radius of these coils.
Formula used:
$1)B=\dfrac{8}{5\sqrt{5}}\dfrac{{{\mu }_{0}}nI}{R}$
$2)B={{B}_{H}}\tan \theta$

Complete answer:
A Helmholtz galvanometer is a device which generates a nearly uniform magnetic field, with the help of two current carrying coils, placed on the same axis.
Magnetic field generated between the current carrying coils is given by
$B=\dfrac{8}{5\sqrt{5}}\dfrac{{{\mu }_{0}}nI}{R}$
where
$B$ is the magnetic field generated by Helmholtz galvanometer
${{\mu }_{0}}$ is the permittivity of free space
$n$ is the equal number of turns in both coils
$I$ is the current flowing the coils
$R$ is the equal radius of both coils
Let this be equation 1.
Magnetic field generated inside these coils is also proportional to the angle of deflection in the galvanometer. If ${{B}_{H}}$ is the horizontal component of the earth’s magnetic field and if $\theta$ is the deflection in galvanometer, magnetic field generated is given by
$B={{B}_{H}}\tan \theta$
where
$B$ is the magnetic field generated by Helmholtz galvanometer
${{B}_{H}}$ is the horizontal component of earth’s magnetic field
$\theta$ is the angle of deflection in the galvanometer
Let this be equation 2.
Combining both equation 1 and equation 2, we have
${{B}_{H}}\tan \theta =\dfrac{8}{5\sqrt{5}}\dfrac{{{\mu }_{0}}nI}{R}$
Let this be equation 3.
Coming to our question, we are given that
$\begin{aligned} & {{B}_{H}}=\dfrac{80}{\pi }A{{m}^{-1}} \\\ & \theta =45{}^\circ \\\ & n=70 \\\ & R=\dfrac{11}{\sqrt{5}}cm=\dfrac{11}{\sqrt{5}}\times {{10}^{-2}}m \\\ \end{aligned}$
Also, we know that
${{\mu }_{0}}=4\pi \times {{10}^{-7}}$
We are required to find the value of current flowing through the coils of Helmholtz galvanometer.
Rearranging equation 3, we have
${{B}_{H}}\tan \theta =\dfrac{8}{5\sqrt{5}}\dfrac{{{\mu }_{0}}nI}{R}\Rightarrow I=\dfrac{{{B}_{H}}\tan \theta \times R\times 5\sqrt{5}}{8\times {{\mu }_{0}}\times n}$
Substituting the given values in the above expression, we have
$I=\dfrac{{{B}_{H}}\tan \theta \times R\times 5\sqrt{5}}{8\times {{\mu }_{0}}\times n}=\dfrac{\dfrac{80}{\pi }A{{m}^{-1}}\times \tan (45{}^\circ )\times \left( \dfrac{11}{\sqrt{5}}\times {{10}^{-2}} \right)m\times 5\sqrt{5}}{8\times 4\pi \times {{10}^{-7}}\times 70}$
On further simplification, we have,
$I=\dfrac{\dfrac{80}{\pi }\times 1\times \left( 11\times {{10}^{-2}} \right)\times 5}{8\times 4\pi \times {{10}^{-7}}\times 70}A=\dfrac{80\times 11\times 5\times {{10}^{-2}}}{8\times 4\times {{(3.14)}^{2}}\times 70\times {{10}^{-7}}}=\dfrac{4400\times {{10}^{5}}}{22085.504}=1.9\times {{10}^{4}}A$
Converting this to $mA$, we have
$I=1.9\times {{10}^{4}}A=1.9\times {{10}^{7}}mA$
Therefore, the current flowing through the coils of the given Helmholtz galvanometer is equal to $1.9\times {{10}^{7}}mA$.

So, the correct answer is “Option A”.

Note:
Students need to be thorough with conversion formulas. Conversion formulas used in this solution are given as follows:
$\begin{aligned} & {{10}^{2}}cm=1m \\\ & {{10}^{3}}mA=1A \\\ \end{aligned}$
They should also try to remember the value of $\tan 45{}^\circ$and the value of ${{\mu }_{0}}$(permittivity of free space). They are:
$\begin{aligned} & \tan 45{}^\circ =1 \\\ & {{\mu }_{0}}=4\pi \times {{10}^{-7}} \\\ \end{aligned}$