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Moving Charges & Magnetism (Part 3)
1. Torque on a Rectangular Current Loop
When a current-carrying coil is placed in a uniform magnetic field, it experiences a torque (turning effect), but no net force.
Derivation of Torque ($\tau$)
Consider a loop of Area $A$ carrying current $I$ in magnetic field $B$.
Force on arms: $F = IlB$.
Torque = Force $\times$ Perpendicular distance
$\tau = F \times (b \sin \theta)$
$$\vec{\tau} = \vec{m} \times \vec{B}$$
Where:
• $\vec{m} = NIA$ (Magnetic Dipole Moment)
• $\theta$: Angle between Normal to the coil and Magnetic Field.
2. The Magnetic Dipole
Every current-carrying loop acts like a magnetic dipole (a tiny bar magnet).
- Magnetic Moment ($\vec{m}$): Product of current and area.
$$m = I \times A$$
Direction: Given by Right Hand Thumb Rule. - Comparison with Electrostatics:
Electric Dipole: $\vec{\tau} = \vec{p} \times \vec{E}$
Magnetic Dipole: $\vec{\tau} = \vec{m} \times \vec{B}$
3. Magnetic Moment of Revolving Electron
An electron revolving around the nucleus constitutes a current, thus creating a magnetic moment.
$$\mu_B = \frac{e h}{4\pi m_e} \approx 9.27 \times 10^{-24} \text{ A m}^2$$
Moving Charges & Magnetism (Part 3)
1. Moving Coil Galvanometer
A device used to detect small electric currents.
Principle: A current-carrying coil placed in a magnetic field experiences a torque.
The pole pieces are made cylindrical (concave) and a soft iron core is placed inside. This ensures that:
1. Magnetic field lines are always radial.
2. Angle between Area Vector and B-field is always $90^\circ$ ($\sin 90^\circ = 1$).
3. Torque becomes maximum and linear ($\tau = NIAB$).
2. Theory & Working
Deflecting Torque: $\tau_d = NIAB$
Restoring Torque: $\tau_r = k \phi$ (where $k$ is torsion constant, $\phi$ is deflection).
At equilibrium: $\tau_d = \tau_r$
$$\phi = \left( \frac{NAB}{k} \right) I$$
This implies Deflection $\phi \propto$ Current $I$ (Linear Scale).
3. Sensitivity of Galvanometer
| Current Sensitivity ($I_s$) | Deflection per unit current. $$I_s = \frac{\phi}{I} = \frac{NAB}{k}$$ |
| Voltage Sensitivity ($V_s$) | Deflection per unit voltage. $$V_s = \frac{\phi}{V} = \frac{\phi}{IR} = \frac{NAB}{kR}$$ |
Moving Charges & Magnetism (Part 3)
1. Conversion to Ammeter
An Ammeter is a low-resistance device used to measure current in series.
Method: Connect a very low resistance called Shunt ($S$) in Parallel with the Galvanometer.
$$S = \left( \frac{I_g}{I - I_g} \right) G$$
• $G$: Galvanometer Resistance
• $I_g$: Full scale deflection current
• $I$: Range of Ammeter required
2. Conversion to Voltmeter
A Voltmeter is a high-resistance device used to measure potential difference in parallel.
Method: Connect a high resistance ($R$) in Series with the Galvanometer.
$$R = \frac{V}{I_g} - G$$
• $V$: Range of Voltmeter required
Question Bank (Part 3)
Section A: MCQs (1 Mark)
Q1. To increase the current sensitivity of a moving coil galvanometer, we should:
(a) Decrease number of turns (b) Increase magnetic field (c) Decrease area (d) Increase torsion constant $k$
Ans: (b) Increase magnetic field (Since $I_s = NAB/k$)
Q2. An ammeter is formed by connecting a shunt parallel to a galvanometer. The resistance of ammeter is:
(a) Infinite (b) Zero (c) Less than shunt (d) More than galvanometer
Ans: (c) Less than shunt (Parallel combination is always lower than smallest resistor).
Section B: Numericals (3 Marks)
Q3. A galvanometer having a coil resistance of $60 \Omega$ shows full scale deflection when a current of 1.0 A passes through it. It can be converted into an ammeter to read currents up to 5.0 A by:
Solution:
Given: $G = 60 \Omega$, $I_g = 1.0$ A, $I = 5.0$ A.
Formula: $S = \frac{I_g G}{I - I_g}$
$S = \frac{1 \times 60}{5 - 1} = \frac{60}{4} = 15 \Omega$.
Ans: Connect $15 \Omega$ in parallel.
Q4. A galvanometer coil has a resistance of $12 \Omega$ and the meter shows full scale deflection for a current of 3 mA. How will you convert the meter into a voltmeter of range 0 to 18 V?
Solution:
Given: $G = 12 \Omega$, $I_g = 3 \text{ mA} = 3 \times 10^{-3}$ A, $V = 18$ V.
Formula: $R = \frac{V}{I_g} - G$
$R = \frac{18}{3 \times 10^{-3}} - 12 = 6000 - 12 = 5988 \Omega$.
Ans: Connect $5988 \Omega$ in series.
Section C: Long Answer (5 Marks)
Q5. Explain the principle and working of a Moving Coil Galvanometer. Why is a radial magnetic field used?
(Refer to Vol 2 of this post).
Q6. Derive the expression for the torque acting on a rectangular current loop placed in a uniform magnetic field.
(Refer to Vol 1 of this post).
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