Unit 03: Magnetic Effects of Current • Chapter 04
Torque on Current Loop & Moving Coil Galvanometer
Magnetic dipole torque, radial magnetic field mechanics, current sensitivity, and conversion to ammeter & voltmeter.
Socho ek revolving door ko: agar aap uske dono opposite ends par equal aur opposite push lagate hain, toh door kahin aage-peeche move nahi hota, balki apni axis par tezi se ghoomne lagta hai (Pure Torque). Jab ek current loop ko magnetic field me rakhte hain, toh uske opposite arms par equal aur opposite forces lagte hain jo ek couple form karte hain. Moving Coil Galvanometer (MCG) isi mechanical torque ko use karta hai. Par agar magnetic field flat hoti, toh coil ke ghoomte hi torque badal jata ($\tau \propto \sin\theta$). Isliye physics engineers ne Cylindrical Soft Iron Core aur Concave Magnet Poles invent kiye, jo ek Radial Magnetic Field banate hain jahan har orientation me plane of coil field ke parallel rehta hai ($\theta = 90^\circ$), aur deflection hamesha current ke directly proportional rehta hai!
1. Torque on Current Loop
Deflecting torque experienced by a coil of $N$ turns and area $A$[cite: 14]:
Where $\theta$ is the angle between magnetic moment $\vec{M}$ (normal to area) and $\vec{B}$[cite: 14]. In a radial field, $\theta = 90^\circ$ always, so $\tau_{\max} = N I A B$[cite: 14].
2. Galvanometer Equilibrium
Deflecting torque balances spring restoring torque ($k\phi$)[cite: 14]:
Where $k$ is torsional restoring couple per unit twist of the phosphor-bronze suspension spring[cite: 14]. Scale is strictly linear ($\phi \propto I$)[cite: 14].
Working Principle & Radial Field Mechanics of Moving Coil Galvanometer[cite: 14]
1. A rectangular coil of $N$ turns, length $l$, and breadth $b$ (area $A = l \times b$) is suspended in a radial magnetic field between concave permanent magnetic poles with a soft iron cylinder core[cite: 14].
2. Due to the cylindrical soft iron core and concave pole pieces, magnetic field lines are directed along radii of the cylinder[cite: 14]. Consequently, the plane of the coil remains parallel to magnetic field lines in all angular positions[cite: 14].
3. Angle between magnetic moment $\vec{M}$ and $\vec{B}$ is strictly $\theta = 90^\circ$ ($\sin 90^\circ = 1$)[cite: 14]. Deflecting torque on the coil is always maximum[cite: 14]:
4. As the coil rotates through angle $\phi$, the suspension spring is twisted, developing an opposing restoring torque[cite: 14]:
5. In steady deflection equilibrium, $\tau_{\text{def}} = \tau_{\text{rest}}$[cite: 14]:
Figure of Merit ($G_c$): Current required to produce unit deflection: $G_c = \frac{I}{\phi} = \frac{k}{N A B}$[cite: 17].
Conversion of Galvanometer into Ammeter[cite: 14]
A galvanometer is converted into an ammeter of range $0$ to $I$ ($I > I_g$) by connecting a small resistance (Shunt $S$) in parallel with it[cite: 14].
Since $G$ and $S$ are in parallel, potential drop across both is equal[cite: 14]:
Equivalent resistance of ammeter: $R_A = \frac{G S}{G + S} \approx S \approx 0$ (Ideal ammeter has zero resistance)[cite: 14].
Conversion of Galvanometer into Voltmeter[cite: 14]
A galvanometer is converted into a voltmeter of range $0$ to $V$ by connecting a very high resistance $R$ in series with it[cite: 14].
Total voltage $V$ is distributed across series combination[cite: 14]:
Equivalent resistance of voltmeter: $R_V = G + R \to \infty$ (Ideal voltmeter has infinite resistance)[cite: 14].
Trap 1: "Increasing Current Sensitivity Increases Voltage Sensitivity" Myth
*Board Question:* If the number of turns $N$ of a galvanometer coil is doubled, does voltage sensitivity double?
*Correct Answer:* NO!
Current sensitivity: $I_s = \frac{\phi}{I} = \frac{NAB}{k} \propto N$[cite: 14]. (Doubles)
Voltage sensitivity: $V_s = \frac{\phi}{V} = \frac{NAB}{k R} \propto \frac{N}{R}$.
When turns $N$ double, length of wire doubles, so resistance $R$ also doubles ($R \propto N$). Therefore, Voltage Sensitivity remains completely UNCHANGED!
Trap 2: The Two Functions of Cylindrical Soft Iron Core
*Board Question:* Why is a soft iron core placed inside a galvanometer coil?
1. High relative magnetic permeability ($\mu_r \gg 1$) draws magnetic lines into itself, drastically increasing magnetic field strength $B$ and sensitivity.
2. Symmetrical cylindrical geometry makes magnetic field lines strictly RADIAL, keeping deflecting torque maximum ($\tau = NIAB$) and independent of coil orientation.
Trap 3: Why Galvanometer Cannot Be Used Directly as Ammeter
1. An unmodified galvanometer is extremely sensitive; full scale deflection occurs at micro-amperes ($\mu\text{A}$), so large currents burn the coil.
2. Galvanometer coil has non-negligible resistance $G$. Inserting it in series in a circuit changes the total resistance and reduces the current being measured[cite: 14]. Shunt $S$ bypasses 99% of current and reduces equivalent resistance to near zero[cite: 14].
Trap 4: Phosphor-Bronze Suspension Wire
Phosphor-bronze alloy is chosen exclusively because it has an extremely small torsional couple per unit twist ($k$), high tensile strength, and negligible elastic after-effects, producing large deflection for small current ($\phi \propto 1/k$).
Problem: A galvanometer coil has a resistance of $12\text{ }\Omega$ and the meter shows full scale deflection for a current of $3\text{ mA}$. How will you convert the meter into a voltmeter of range $0$ to $18\text{ V}$[cite: 14]?
Solution:
1. Given: Galvanometer resistance $G = 12\text{ }\Omega$, full scale current $I_g = 3\text{ mA} = 3 \times 10^{-3}\text{ A}$, target range $V = 18\text{ V}$[cite: 14].
2. To convert into a voltmeter, a high resistance $R$ must be connected in series[cite: 14]:
Verdict: Connect a resistance of $5988\text{ }\Omega$ in series with the galvanometer coil[cite: 14]. Total voltmeter resistance becomes $6000\text{ }\Omega$.
Problem: A galvanometer has a resistance of $50\text{ }\Omega$ and gives full-scale deflection with $0.5\text{ mA}$ current. A shunt of $0.05\text{ }\Omega$ is connected to convert it into an ammeter. If the meter now reads $0.4\text{ A}$, find the actual current passing through the galvanometer coil and through the shunt.
Solution:
1. Given: $G = 50\text{ }\Omega$, $S = 0.05\text{ }\Omega$, total entering circuit current $I = 0.4\text{ A}$.
2. Shunt and galvanometer are in parallel, sharing total current $I$ inversely to resistance:
3. Current through the shunt:
Observation: Out of $400\text{ mA}$ entering current, $399.6\text{ mA}$ (99.9%) flows safely through the shunt, protecting the delicate coil!
MCG Linear Law
$\phi = \left(\frac{NAB}{k}\right) I$[cite: 14]
Ammeter Shunt
$S = \frac{I_g G}{I - I_g}$[cite: 14]
Voltmeter Series R
$R = \frac{V}{I_g} - G$[cite: 14]
Radial Field
$\sin\theta = 1$ always[cite: 14]