Physics, asked by pankajpankajg4915, 1 year ago

Explain the working of a moving coil galvanometer is based on this principle of torque

Answers

Answered by sarika65
0
Let PQRS be a single turn of the coil. A current I flows through the coil. In a radial magnetic field, the plane of the coil is always parallel to the magnetic field. Hence the sides QR and SP are always parallel to the field. So, they do not experience any force. The sides PQ and RS are always perpendicular to the field.

PQ = RS = l, length of the coil and PS = QR = b, breadth of the coil. Force on PQ, F = BI (PQ) = BIl. According to Fleming’s left-hand rule, this force is normal to the plane of the coil and acts outwards.



Force on RS, F = BI (RS) = BIl. This force is normal to the plane of the coil and acts inwards. These two equal, oppositely directed parallel forces having different lines of action constitute a couple and deflect the coil. If there are n turns in the coil, the moment of the deflecting couple = n BIl – b

Hence the moment of the deflecting couple = nBIA

When the coil deflects, the suspension wire is twisted. On account of elasticity, a restoring couple is set up in the wire. This couple is proportional to the twist. If θ is the angular twist, then, the moment of the restoring couple = Cθ, where C is the restoring couple per unit twist. Atequilibrium, deflecting couple = restoring couple nBIA = Cθ

Hence we can write, nBIA = Cθ

I = (C / nBA) × θ where C is the torsional constant of the spring; i.e. the restoring torque per unit twist. The deflection θ is indicated on the scale by a pointer attached to the spring.

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