Question

Derive an expression for the torque acting on a loop of

N

turns, area

A

, carrying current

I

, when held in a magnetic field

B

. With the help of a circuit diagram, show how a moving coil galvanometer can be converted into an ammeter of given range. Write the necessary mathematical formula.

Accepted Answer

Let $I =$ current flowing through coil $B$

$a, b$ are sides of coil $P Q R S$

The area of the coil, $A = a b$

The number of turns, $= N$

The angle between the direction of $\vec{B}$ and normal to the plane of the coil, $= θ$

According to Fleming’s left-hand rule, the magnetic forces on sides $P S$ and $Q R$ are equal, opposite and collinear, so their resultant is zero.

The side $P Q$ experience a normal inward force equal to $I b B$ while the side $R S$ experiences an equal normal outward force. These two forces form a couple which exerts a torque given by,

$τ = Force \times pependicular distance$

$= N I b B \times a \sin θ = N I B A \sin 90 ° = N I B A$

Here, $θ = 90 °$ , because the normal to the plane of the coil remains perpendicular to the field $\vec{B}$ in all positions.

A galvanometer can be converted into an ammeter by connecting a low resistance called shunt in parallel to the galvanometer as shown in the figure below.

The resistance of galvanometer, $= G$

Current in galvanometer for full-scale deflection, $I_{g}$

The range of ammeter, $0$ to $I$

Let the required shunt resistance is $S$ and the current through the shunt is $I - I_{g}$

The shunt resistance is connected in parallel with the galvanometer. Hence,
P.D. across the galvanometer $=$ P.D. across the shunt

$I_{g} G = (I - I_{g}) S$

$\Rightarrow S = \frac{I_{g}}{I - I_{g}} \times G$

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