Question
Explain the construction and working of the Moving coil Galvanometer.
✓
Answer
Construction:
i. M.C.G. consists of a coil of several turns mounted (suspended or pivoted) in such a way that it can freely rotate about a fixed axis, in a radial uniform magnetic field.
ii. A soft iron cylindrical core makes the field radial and strong.
Moving coil galvanometer
Working:
i. The coil rotates due to a torque acting on it as the current flows through it. The torque acting on the current-carrying coil is $\tau=$ NIAB $\sin \theta$.
Here $\theta=90^{\circ}$ as the field is radial.
$\therefore \tau= NIAB$
where $A$ is the area of the coil, $B$ the strength of the magnetic field, $N$ the number of turns of the coil, and I the current in the coil.
ii. This torque is counter balanced by a torque due to a spring fitted at the bottom so that a fixed steady current I in the coil produces a steady angular deflection $\Phi$.
iii. The larger the current is, the larger is the deflection and the larger is the torque due to the spring. If the deflection is $\Phi$, the restoring torque due to the spring is equal to $K \Phi$ where $K$ is the torsional constant of the spring. Thus, $K \Phi=N I A B$, and the deflection $\Phi=$ $\left(\frac{ NAB }{ K }\right) I$
This means the deflection $\Phi$ is proportional to the current I i.e., $\Phi \propto I$.
i. M.C.G. consists of a coil of several turns mounted (suspended or pivoted) in such a way that it can freely rotate about a fixed axis, in a radial uniform magnetic field.
ii. A soft iron cylindrical core makes the field radial and strong.
Moving coil galvanometer
Working:
i. The coil rotates due to a torque acting on it as the current flows through it. The torque acting on the current-carrying coil is $\tau=$ NIAB $\sin \theta$.
Here $\theta=90^{\circ}$ as the field is radial.
$\therefore \tau= NIAB$
where $A$ is the area of the coil, $B$ the strength of the magnetic field, $N$ the number of turns of the coil, and I the current in the coil.
ii. This torque is counter balanced by a torque due to a spring fitted at the bottom so that a fixed steady current I in the coil produces a steady angular deflection $\Phi$.
iii. The larger the current is, the larger is the deflection and the larger is the torque due to the spring. If the deflection is $\Phi$, the restoring torque due to the spring is equal to $K \Phi$ where $K$ is the torsional constant of the spring. Thus, $K \Phi=N I A B$, and the deflection $\Phi=$ $\left(\frac{ NAB }{ K }\right) I$
This means the deflection $\Phi$ is proportional to the current I i.e., $\Phi \propto I$.
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