Question

(a) Why is the magnetic field radial in a moving coil galvanometer? Explain how it is achieved
(b) A galvanometer of resistance ‘$G$’ can be converted into a voltmeter of range ($0 - V$) volts by connecting a resistance ‘$R$’ in series with it. How much resistance will be required to change its range from $0$ to $\dfrac{V}{2}$?

Answer 1

A galvanometer is a device used in-order to detect current in the circuit which consists of a current carrying coil that rotates in order to produce the magnetic field because of the torque acting on it. The answer to why the magnetic field produced is radial is related to the construction and the corresponding working of this device and the placement of the cylindrical iron core between the magnets. The formula for the equation of a galvanometer converted voltmeter is applied to determine the resistance for a change in voltage over a particular range.

Formula used:
The high series resistance need to convert a galvanometer into a voltmeter is given by the equation:
$R = \dfrac{V}{{{I_g}}} - G$
Where, $V$ is the voltage, ${I_g}$ is the current for which the galvanometer gives full scale deflection and $G$ is the resistance of the galvanometer.
The equation for the torque is given by:
$\tau = NIbB \times a\sin \theta$
Where, $I$ is the current through the coil, $A = ab$ is the area of the coil, $\vec B$ is the magnetic field and $N$ is the number of turns.

Complete step by step answer:
We are asked to determine why a galvanometer consists of a radial field. To know this we first need to learn the construction of the galvanometer.

(a) The galvanometer consists of a rectangular coil mounted to two hair springs on either side of the coil. The coil is wound with insulated copper wires. The hair springs are responsible for the motion of the coil which ensures that it provides the access for the coil to rotate. These springs are responsible for the production of restoring torque and this torque is responsible for the deflection in the galvanometer.

The equation for the torque is given by:
$\tau = NIbB \times a\sin \theta$-----($1$)
The main reason for the magnetic field to be produced is because of the soft iron core cylinder which is symmetrically placed between the poles of a strong permanent horseshoe magnet. This makes the magnetic lines of force point along the radii of the circle. We take into consideration the area of cross-section of the top of the iron cylinder and when we view from its top view the lines of force seem to pass through the radius of the circle and hence it is known as radial field. The plane of the coil rotates in such a way that the field always remains parallel to it from all the positions.

Now coming to the explanation of the need for this field to be radial. The principle of the galvanometer is to produce deflection even when a small current is passed through it which means it must be sensitive. This also means that the current must be directly proportional to the deflection produced in the galvanometer, that is, with an increase in current the deflection is increased or produced. The equation for the deflection is given by the relation:
$\phi = \dfrac{{NIAB \times \sin \theta }}{k}$
Where, $k$ is the torsion constant.

Here, $\theta$ is the angle between the magnetic field vector and the normal to the plane. Our aim is to make the current directly proportional to the deflection and this can only be possible if we get rid of the sine term, that is, only if $\sin \theta = 1$ so $\theta$ must be ${90^ \circ }$ as all the other terms are constants. So in-order to achieve this condition in accordance with the concept of a galvanometer we produce a radial field.

(b) It is given in the question that in-order to convert a galvanometer into a voltmeter we must connect a series resistance to it. A galvanometer is designed to measure only small amounts of current and hence a high series resistor is used to limit this current to the current value suitable for a galvanometer.

The range of the voltmeter is given to be $V$ initially and then changed to the range $0$ to $\dfrac{V}{2}$. We are asked to find out the new resistance after the change in voltage range. We know that this resistance is dependent on the current which will produce full scale deflection. Hence, the relation for the resistance is given by the equation:
$R = \dfrac{V}{{{I_g}}} - G$
By the taking the LCM of the above equation we get:
$R = \dfrac{{V - G{I_g}}}{{{I_g}}}$
By rearranging the terms let us determine the equation for the current:
$\Rightarrow R{I_g} = V - G{I_g}$
$\Rightarrow R{I_g} + G{I_g} = V$
We take out the common terms:
$\Rightarrow {I_g}(R + G) = V$
$\Rightarrow {I_g} = \dfrac{V}{{R + G}}$ -----($2$)

As per the question the galvanometer resistance and the current is not varied and changes to the voltage range is made. This change in voltage is given to be ranging from $0$ to $\dfrac{V}{2}$. Let ${R_1}$ be the new resistance of the voltmeter when the potential difference is changed. Hence we get another equation for current from equation ($2$) to be:
${I_g} = \dfrac{{\left( {\dfrac{V}{2}} \right)}}{{{R_1} + G}}$
$\Rightarrow {I_g} = \dfrac{V}{{2({R_1} + G)}}$-----($3$)
Since the equations ($2$) and ($3$) are identical to each other they are equated and we get:
$\Rightarrow \dfrac{V}{{R + G}} = \dfrac{V}{{2({R_1} + G)}}$
The common terms are cancelled out to get:
$\Rightarrow \dfrac{1}{{R + G}} = \dfrac{1}{{2({R_1} + G)}}$

By cross multiplying the terms and solving out the equation we get the new changed value of resistance.
$\Rightarrow 2({R_1} + G) = R + G$
$\Rightarrow 2{R_1} + 2G = R + G$
$\Rightarrow 2{R_1} = R + G - 2G$
$\Rightarrow 2{R_1} = R - G$
$\therefore {R_1} = \dfrac{{R - G}}{2}$
This is the new value of resistance that must be applied for a voltmeter of range $0$ to $\dfrac{V}{2}$. Therefore, the amount of resistance required to change the range of the voltmeter’s range is $\dfrac{{R - G}}{2}$.

Additional information: The principle of a moving coil galvanometer is to produce a current dependent torque which tends to automatically rotate the coil of the galvanometer due to the force of the magnetic field produced. A voltmeter is a device that is used to measure the potential difference between two points in a circuit and hence it is connected in parallel to the circuit element for which the voltage is to be measured.

Due to this connection by Kirchhoff's law for the splitting of current in the junction the current is said to split up and a small part of the current goes to the voltmeter as well decreasing the current across the circuit element. This results in a subsequent decrease in potential difference and hence in-order to avoid this there is a need to use a resistor with a high resistance in the voltmeter set up. This is the reason for using a voltmeter designed to offer a high value of resistance.

Note: This explanation for why a radial field is to be produced can be made in an alternative way as well, that is, in terms of the concept of torque. The purpose of a galvanometer is to produce a deflection due to the production of current which is torque dependent. In-order to produce this torque a couple force needs to be produced (because torque is dependent on the force and the perpendicular distance between them). This is only possible when a radial field is produced, that is, the plane of the coil must parallel to the field. The aim here is to produce maximum torque and as per equation ($1$) this is only possible when $\theta = {90^ \circ }$ from the equation and this happens only when there is a radial field.