Question

The radius of each coil of a Helmholtz galvanometer is $0.1m$ and the number of turns in each is $25$ . When a current is passed in it then the deflection of the magnetic needle is observed as ${45^ \circ }$ . If the horizontal component of earth’s magnetic field is $0.314 \times {10^{ - 4}}T$ then the value of current will be
(A) $0.14{\text{A}}$
(B) $0.28{\text{A}}$
(C) $0.42{\text{A}}$
(D) $0.07{\text{A}}$

Answer 1

To solve this question, we need to use the formula for the magnetic field produced due to a current carrying coil at its centre. Then we need to use the value of the deflection of the needle to deduce a relation between this magnetic field and the horizontal component of earth’s magnetic field.

Formula used:
The formula used to solve this question is given by
$\Rightarrow B = \dfrac{{8{\mu _0}nI}}{{5\sqrt 5 R}}$ , here $B$ is the magnetic field produced between the two Helmholtz coils, $n$ and $R$ are respectively the number of turns and radius of each coil, $I$ is the current in each coil, and ${\mu _0}$ is the magnetic permeability in vacuum.

Complete step by step solution:
As there are two magnetic fields involved; one due to the current carrying coils, and the other due to the earth’s magnetic field, so the resultant magnetic field is the vector sum of the horizontal component of the earth’s magnetic field, and the field produced by the Helmholtz coils. We know that the total magnetic field produced by the coils of a Helmholtz galvanometer is given by
$B = \dfrac{{8{\mu _0}nI}}{{5\sqrt 5 R}}$ ……………...(1)
The vector diagram for the resultant magnetic field can be represented as below

We know that a magnetic field points in the direction of the resultant magnetic field, ${B_N}$ . According to the question, the deflection of the magnetic needle is equal to ${45^ \circ }$ . So we have from the above figure
$\theta = {45^ \circ }$
Also from the above figure
$\Rightarrow \tan \theta = \dfrac{B}{{{B_H}}}$
$\Rightarrow \tan {45^ \circ } = \dfrac{B}{{{B_H}}}$
We know that $\tan {45^ \circ } = 1$ . So we get
$\Rightarrow \dfrac{B}{{{B_H}}} = 1$
$\Rightarrow B = {B_H}$
From (1)
$\Rightarrow \dfrac{{8{\mu _0}nI}}{{5\sqrt 5 R}} = {B_H}$
So we get the current in the coil as
$\Rightarrow I = \dfrac{{5\sqrt 5 R{B_H}}}{{8{\mu _0}n}}$ …………….(2)
Now, according to the question we have the radius of each coil $r = 0.1m$ , the number of turns $n = 25$ , the value of the horizontal component of the earth’s magnetic field ${B_H} = 0.314 \times {10^{ - 4}}T$ and as we know that ${\mu _0} = 4\pi \times {10^{ - 7}}$ . Substituting these in (2) we get
$\Rightarrow I = \dfrac{{5\sqrt 5 \times 0.1 \times 0.314 \times {{10}^{ - 4}}}}{{8 \times 4\pi \times {{10}^{ - 7}} \times 25}}$
On solving, we finally get the current as,
$\Rightarrow I = 0.139{\text{A}} \approx {\text{0}}{\text{.14A}}$
Thus we get the value of the current equal to ${\text{0}}{\text{.14A}}$ .
So the correct answer is option A.

Note:
We should not use the formula for the magnetic field which we have used in this solution as the value of the magnetic field on the common axis of the two coils at their mid point. And we also should not use the formula for the magnetic field at the centre of a circular current carrying loop, as it is due to a single coil.