The Moving-Coil Galvanometer

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. In the last lesson, the torque on a coil in a uniform field depended on its angle, so it kept changing as the coil turned. But look at any galvanometer, the meter that detects tiny currents. Its scale is evenly marked: twice the current, twice the deflection. How can that be, if the torque keeps changing? By the end of this video, you will know. This comes up often in JEE, so let's begin.

0:32The radial field: same torque at every position

So, here is the heart of a moving-coil galvanometer: a rectangular coil of capital N turns, each of area capital A, carrying a current, capital I, and free to turn between the poles of a magnet. Okay? Now, suppose the field between the poles were uniform, like in the last lesson. As the coil turns, does the torque stay the same? No, it does not, and that is our problem. The torque depends on the angle between the coil's normal and the field, so the deflection would not be proportional to the current. With me? So the galvanometer uses concave pole pieces, hollowed into a curve, and a soft-iron cylinder inside the coil. The soft iron concentrates the field lines, so the field in the gap is stronger, and radial, like the spokes of a wheel. Right? Now see the beauty of it. Whatever the coil's position, its plane contains B vector; the small arrow marks a vector. So the normal is always at ninety degrees to the field, and the torque, tau, the Greek letter, is NIAB at every position. Clear?

1:34Balancing the torques: a linear scale

Next, a spring, or a suspension fibre, holds the coil back. When the coil turns through an angle theta, the Greek letter, the spring gives a restoring torque k theta, where k is the torsion constant. Okay? The magnetic torque turns the coil, and the spring's torque grows as it turns. So, where does the coil finally stop? Yes, exactly, it stops where the two torques balance. At equilibrium, NIAB equals k theta. Got it? So theta equals NAB over k, times I, the current. This sign, like a stretched alpha, means proportional to. So theta is proportional to the current, and our puzzle is solved: thanks to the radial field, the scale is linear. Right?

2:19Sensitivity and figure of merit

Now, how sensitive is it? The current sensitivity is the deflection per unit current, theta over I, which equals NAB over k. With me? The voltage sensitivity is theta over capital V, the deflection per unit voltage. The voltage is the current times capital G, the coil's resistance, so G joins k in the denominator, giving NAB over k G, okay? And the figure of merit is the current per unit deflection, I over theta, equal to k over NAB, just the reciprocal of the current sensitivity. Clear?

2:51The trap: double the turns

Now a question that comes up often. We double the number of turns, with the same wire and the same area. N is on top in NAB over k, so the current sensitivity doubles. Easy, isn't it? So the voltage sensitivity also doubles, right? What do you think? No, no, careful, that is exactly the trap! With the same wire, the coil's resistance, G, also doubles, in the same ratio. Okay? So the voltage sensitivity, which goes as N over G, does not change at all. More turns raise the current sensitivity, but not the voltage sensitivity. Got it?

3:30Bringing the pointer to rest

One more practical point. The coil is wound on a metal frame. Why metal, do you think? Because as the frame swings in the field, eddy currents flow in it and damp the motion. So the pointer comes to rest quickly, instead of swinging for a long time.

3:48Example 1: current sensitivity and deflection for 1 mA

Right? Now the first example. Here, a galvanometer coil has one hundred turns of area two square centimetres, in a radial field of zero point one tesla, written capital T, okay? The torsion constant is ten to the power minus five newton metre per radian. Find its current sensitivity, and the deflection for one milliampere, written m A on screen. Pause the video and try it first. Done? Okay, let's check. The centimetre is squared, so ten to the minus two becomes ten to the minus four, and the area is two times ten to the minus four square metre. Right? Now the current sensitivity, NAB over k. One hundred times two times ten to the minus four, times zero point one, all over ten to the minus five. With me? The top gives two times ten to the minus three. Over ten to the minus five, that gives two hundred radians per ampere, the rad A to the power minus one on screen. Okay? For one milliampere, which is ten to the minus three ampere, theta equals two hundred times ten to the minus three, which gives zero point two. So that is zero point two degrees, right? No, careful, it is radians, because the torsion constant was given per radian. Zero point two radian is about eleven point five degrees; the wavy equals sign means approximately, and the small circle means degrees. Clear?

5:13Example 2: sudden charge Q, find the maximum deflection

Now a harder one, with a lovely idea in it. This time, a charge capital Q passes through the coil in a time much shorter than its period. The coil has moment of inertia I c, capital I with a small c below, N turns of area A, field B, and torsion constant k. Find the maximum deflection. Pause the video and try it first. Done? Okay, let's check. The pulse is so short that the coil has hardly moved when it ends. So during the pulse, what does the coil receive? It gets a kick, an angular impulse, exactly! At each instant the torque is NAB times small i, the current at that instant. The tall S-shaped sign is the integral: we add up the torque, times d t, over the pulse. The integral of i, d t, is the total charge, so the angular impulse is NAB times Q, the charge. With me? This impulse becomes the coil's angular momentum. So NABQ equals I c times omega naught, where omega naught, the Greek letter omega with a small zero, is the angular speed it starts with. Okay? Now the coil swings out, and the spring slows it down. At the largest deflection it stops for a moment. So where did its kinetic energy go? It is stored in the twisted spring, exactly! So half times I c times omega naught squared equals half times k times theta max squared, where theta max, with max written small below, is the largest deflection. Clear? Now let us solve for theta max. It equals omega naught times the square root of I c over k; the tick-shaped sign is the square root. From the second line, omega naught is NABQ over I c. Put it in, and theta max equals NABQ, over the square root of k times I c. So the largest swing is proportional to the charge, and that is the whole answer.

7:11Quick revision: radial field, sensitivities, eddy damping

So, quick revision. The radial field makes the torque NIAB everywhere, so theta equals NAB over k, times I, and the scale is linear. The current sensitivity is NAB over k, the voltage sensitivity is NAB over k G, and more turns of the same wire leave the voltage sensitivity unchanged. And eddy currents in the metal frame bring the pointer to rest quickly. Now go and try the practice problems on this page. You can do them, all the best!