Question

A flat circular coil of 200 turns of diameter 25 cm is laid on a horizontal table and connected to a ballistic galvanometer. The complete circuit is having resistance of 800Ω. When the coil is quickly turned over, the spot of light swings to a maximum reading of 30 divisions. When a 0.1$\mu$F capacitor charged to 6 V is discharged through the same ballistic galvanometer, a maximum reading of 20 division is obtained. Calculate the vertical component of the earth's magnetic induction.

Answer 1

3.667 x 10^{-5} T

Answer 2

The problem involves two scenarios with a ballistic galvanometer, which measures the total charge passed through it. The deflection of a ballistic galvanometer is proportional to the charge. Let $k$ be the proportionality constant (galvanometer constant).

Scenario 1: Coil turned over in Earth's magnetic field

1. Initial Magnetic Flux: When the coil is laid on a horizontal table, the vertical component of the Earth's magnetic induction ($B_v$) passes perpendicularly through its area. The initial magnetic flux ($\Phi_i$) through the coil is given by: $\Phi_i = N B_v A$ where $N$ is the number of turns, $B_v$ is the vertical component of the Earth's magnetic induction, and $A$ is the area of the coil.

2. Final Magnetic Flux: When the coil is quickly turned over, the direction of the area vector effectively reverses with respect to the magnetic field. The final magnetic flux ($\Phi_f$) is: $\Phi_f = -N B_v A$

3. Change in Magnetic Flux: The magnitude of the change in magnetic flux ($\Delta \Phi$) is: $\Delta \Phi = |\Phi_f - \Phi_i| = |-N B_v A - N B_v A| = |-2N B_v A| = 2N B_v A$

4. Induced Charge: According to Faraday's law of electromagnetic induction and Ohm's law, the total charge ($Q_1$) that flows through the circuit due to this flux change is: $Q_1 = \frac{\Delta \Phi}{R} = \frac{2N B_v A}{R}$ where $R$ is the total resistance of the circuit.

5. Galvanometer Deflection: This charge causes a deflection of 30 divisions in the ballistic galvanometer. $Q_1 = k \times 30 \quad (Equation \ 1)$

Scenario 2: Capacitor discharge

1. Charge from Capacitor: A capacitor of capacitance $C$ charged to a voltage $V$ stores a charge $Q_2$ given by: $Q_2 = CV$

2. Galvanometer Deflection: When this capacitor is discharged through the same ballistic galvanometer, a deflection of 20 divisions is obtained. $Q_2 = k \times 20 \quad (Equation \ 2)$

Calculations

1. Determine the galvanometer constant ($k$): From Equation 2: $k = \frac{Q_2}{20} = \frac{CV}{20}$

2. Substitute $k$ into Equation 1: $\frac{2N B_v A}{R} = \left(\frac{CV}{20}\right) \times 30$ $\frac{2N B_v A}{R} = \frac{3CV}{2}$

3. Solve for $B_v$: $B_v = \frac{3CVR}{4NA}$

4. Substitute the given values:

   • Number of turns, $N = 200$
   • Diameter of coil, $D = 25 \, \text{cm} = 0.25 \, \text{m}$
   • Radius of coil, $r = D/2 = 0.125 \, \text{m}$
   • Area of coil, $A = \pi r^2 = \pi (0.125)^2 \, \text{m}^2 = \pi (1/8)^2 \, \text{m}^2 = \frac{\pi}{64} \, \text{m}^2$
   • Resistance of the circuit, $R = 800 \, \Omega$
   • Capacitance, $C = 0.1 \, \mu F = 0.1 \times 10^{-6} \, F = 10^{-7} \, F$
   • Voltage, $V = 6 \, V$

   Now, plug these values into the formula for $B_v$: $B_v = \frac{3 \times (10^{-7} \, F) \times (6 \, V) \times (800 \, \Omega)}{4 \times (200) \times (\frac{\pi}{64} \, \text{m}^2)}$ $B_v = \frac{3 \times 10^{-7} \times 6 \times 800}{800 \times \frac{\pi}{64}}$ $B_v = \frac{18 \times 10^{-7}}{\frac{\pi}{64}}$ $B_v = \frac{18 \times 64 \times 10^{-7}}{\pi}$ $B_v = \frac{1152 \times 10^{-7}}{\pi}$ $B_v = \frac{1.152 \times 10^{-4}}{\pi}$

   Using $\pi \approx 3.14159$: $B_v \approx \frac{1.152 \times 10^{-4}}{3.14159}$ $B_v \approx 0.3667 \times 10^{-4} \, \text{T}$ $B_v \approx 3.667 \times 10^{-5} \, \text{T}$

The vertical component of the Earth's magnetic induction is approximately $3.667 \times 10^{-5}$ Tesla.

The final answer is $\boxed{3.667 \times 10^{-5} T}$

Explanation of the solution:

The ballistic galvanometer measures charge. The charge from turning the coil is $Q_1 = \frac{2NBA}{R}$, proportional to deflection $\theta_1$. The charge from capacitor discharge is $Q_2 = CV$, proportional to deflection $\theta_2$.

Thus, $\frac{Q_1}{\theta_1} = \frac{Q_2}{\theta_2}$.

$\frac{2NBA/R}{30} = \frac{CV}{20}$.

Solving for $B$: $B = \frac{3CVR}{4NA}$.

Substitute given values: $N=200$, $A=\pi(0.125)^2$, $R=800$, $C=10^{-7}$, $V=6$.

$B = \frac{3 \times 10^{-7} \times 6 \times 800}{4 \times 200 \times \pi (0.125)^2} = \frac{18 \times 10^{-7}}{\pi/64} = \frac{1152 \times 10^{-7}}{\pi} \approx 3.667 \times 10^{-5} T$.

Answer: $3.667 \times 10^{-5} T$