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A tangent galvanometer has a coil with 5...

A tangent galvanometer has a coil with 50 turns and radius equal to 4 cm . A current of 0.1 A is passing through it. The plane of the coil is set parallel to the earth's magnetic meridian. If the value of the earth's horizontal component of the magnetic field is `7xx10^(-5)` Tesla and `mu_(0)=4pixx10^(-7)Weber//ampxxm` then the deflection in the galvanometer needle will be

A

`45^(@)`

B

`48.2^(@)`

C

`50.7^(@)`

D

`52.7^(@)`

Text Solution

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The correct Answer is:
To solve the problem, we need to find the deflection angle (θ) of the galvanometer needle using the given parameters. We can use the formula for the tangent galvanometer: \[ \tan \theta = \frac{\mu_0 n I}{2 R B_h} \] Where: - \( \mu_0 = 4 \pi \times 10^{-7} \, \text{Weber/Ampere meter} \) - \( n = 50 \) (number of turns) - \( I = 0.1 \, \text{A} \) (current) - \( R = 4 \, \text{cm} = 0.04 \, \text{m} \) (radius) - \( B_h = 7 \times 10^{-5} \, \text{T} \) (horizontal component of the Earth's magnetic field) ### Step 1: Substitute the values into the formula First, we need to substitute the known values into the formula: \[ \tan \theta = \frac{(4 \pi \times 10^{-7}) \times 50 \times 0.1}{2 \times 0.04 \times (7 \times 10^{-5})} \] ### Step 2: Calculate the numerator Calculating the numerator: \[ 4 \pi \times 10^{-7} \times 50 \times 0.1 = 2 \pi \times 10^{-7} \times 5 = 10 \pi \times 10^{-7} \] Using \( \pi \approx 3.14 \): \[ 10 \pi \times 10^{-7} \approx 31.4 \times 10^{-7} = 3.14 \times 10^{-6} \] ### Step 3: Calculate the denominator Now, calculating the denominator: \[ 2 \times 0.04 \times (7 \times 10^{-5}) = 0.08 \times 7 \times 10^{-5} = 5.6 \times 10^{-6} \] ### Step 4: Calculate \( \tan \theta \) Now substituting the values back into the equation: \[ \tan \theta = \frac{3.14 \times 10^{-6}}{5.6 \times 10^{-6}} \approx 0.5607 \] ### Step 5: Find \( \theta \) To find \( \theta \), we take the arctangent: \[ \theta = \tan^{-1}(0.5607) \] Using a calculator, we find: \[ \theta \approx 29.5^\circ \] ### Conclusion The deflection in the galvanometer needle will be approximately \( 29.5^\circ \). ---

To solve the problem, we need to find the deflection angle (θ) of the galvanometer needle using the given parameters. We can use the formula for the tangent galvanometer: \[ \tan \theta = \frac{\mu_0 n I}{2 R B_h} \] Where: - \( \mu_0 = 4 \pi \times 10^{-7} \, \text{Weber/Ampere meter} \) ...
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