CBSE · Class 12 · Physics · Moving Charges and MagnetismExplain the principle, construction, working, and conversion of a Moving Coil Galvanometer into an Ammeter with necessary mathematical derivations.
Principle of Moving Coil Galvanometer
\nA moving coil galvanometer works on the principle that a current-carrying coil placed in a uniform magnetic field experiences a magnetic torque which tends to rotate it, and the deflection produced is directly proportional to the current flowing through the coil.
Construction
- Coil: It consists of a rectangular coil of many turns of insulated copper wire wound on a light non-magnetic frame.
- Magnetic Field: A strong permanent magnet with cylindrical pole pieces provides a radial magnetic field, ensuring that the plane of the coil is always parallel to the magnetic field lines in all orientations.
- Soft Iron Core: A cylindrical soft iron core is placed symmetrically inside the coil to make the field radial and increase the strength of the magnetic field.
- Suspension/Pivot: The coil is suspended by a phosphor-bronze strip from a movable torsion head, which also acts as one terminal.
Working
\nWhen an electric current $I$ flows through the coil, a magnetic torque acts on it. The deflecting torque $\tau$ is given by: $$\tau = NIAB$$\nwhere $N$ is the number of turns, $I$ is the current, $A$ is the area of the coil, and $B$ is the magnetic field strength. \nDue to this torque, the coil rotates, causing a twist in the suspension strip. This creates a restoring torque $\tau_r$ proportional to the angle of twist $\alpha$: $$\tau_r = k\alpha$$\nwhere $k$ is the torsional constant of the suspension fiber. \nAt equilibrium, the deflecting torque balances the restoring torque: $$NIAB = k\alpha$$ $$I = \left(\frac{k}{NAB}\right)\alpha$$\nSince $k, N, A,$ and $B$ are constants, we can write: $$I = G\alpha \implies \alpha \propto I$$\nThus, the deflection is directly proportional to the current.
Conversion of Galvanometer into an Ammeter
\nA galvanometer is a very sensitive device and cannot measure high currents directly. It can be converted into an ammeter by connecting a very small resistance, known as a shunt resistance ($S$), in parallel with the galvanometer. \nLet:
- $G$ = Resistance of the galvanometer
- $I$ = Total current to be measured
- $I_g$ = Maximum current that can pass through the galvanometer (for full-scale deflection)
- $S$ = Shunt resistance connected in parallel \nThe current flowing through the shunt is $(I - I_g)$. Since the galvanometer and the shunt are in parallel, the potential difference across them is equal: $$I_g \cdot G = (I - I_g) \cdot S$$ $$S = \frac{I_g \cdot G}{I - I_g}$| \nThis formula gives the exact value of the shunt resistance required to convert a galvanometer into an ammeter of a desired range.