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Step-by-Step Solution
Step 1: Identify the Relevant Formula
A moving coil galvanometer’s deflection is given by the balance of the magnetic torque and the restoring torque of the torsion band:
$$ N \, I \, A \, B = K \, \theta $$
where
$N$ is the number of turns of the coil,
$I$ is the current through the coil,
$A$ is the area of the coil,
$B$ is the magnetic field,
$K$ is the torsion constant of the suspension,
$\theta$ is the angle of deflection in radians.
Step 2: List the Given Quantities
From the question:
Number of turns, $N = 175$
Current, $I = 1\,\text{mA} = 1 \times 10^{-3}\,\text{A}$
Area, $A = 1\,\text{cm}^{2} = 1 \times 10^{-4}\,\text{m}^{2}$
Torsion constant, $K = 10^{-6}\,\text{N}\cdot\text{m/rad}$
Angle of deflection, $\theta = 1^\circ = \frac{\pi}{180}\,\text{radians}$
Step 3: Substitute the Values into the Formula
Using
$$N \, I \, A \, B = K \, \theta,$$
we solve for $B$:
$$ B = \frac{K \, \theta}{N \, I \, A} .$$
Step 4: Plug in the Numerical Values
Substituting all known values:
$$ B = \frac{(10^{-6}) \times \left(\frac{\pi}{180}\right)}{(175) \times \left(1 \times 10^{-3}\right) \times \left(1 \times 10^{-4}\right)} .$$
Step 5: Simplify the Expression
First, calculate the product in the denominator:
$$ 175 \times 10^{-3} \times 10^{-4} = 175 \times 10^{-7} = 1.75 \times 10^{-5}. $$
Hence,
$$ B = \frac{10^{-6} \times \frac{\pi}{180}}{1.75 \times 10^{-5}}. $$
Step 6: Numerical Evaluation
Compute in steps:
Divide $10^{-6}$ by $1.75 \times 10^{-5}$ to get approximately $0.057$.
Multiply this result by $\frac{\pi}{180} \approx 0.01745$.
The final product is around $0.057 \times 0.01745 \approx 9.9 \times 10^{-4}\,\text{T}.$
So,
$$ B \approx 10^{-3}\,\text{T}. $$
Step 7: Conclude the Result
Thus, the value of the magnetic field that satisfies the given condition is
$$ 10^{-3}\,\text{T}. $$
