The radius of the coil of a Tangent galvanometer, which has 10 turns, is 0.1 m. The current required to produce a deflection of `60^(@) (B_(H)=4xx10^(-5)T)` is

Options

A. (a)3A
B. (b)1.1A
C. (c)2.1A
D. (d)1.5 A

Explanation

To find the current required to produce a deflection of \(60^\circ\) in a tangent galvanometer with 10 turns and a radius of 0.1 m, we can follow these steps:

### Step 1: Understand the relationship between magnetic field and current
The magnetic field \(B\) at the center of a circular coil carrying current \(I\) is given by the formula:
\[ B = \frac{\mu_0 N I}{2R} \]
where:
- \(B\) is the magnetic field,
- \(\mu_0\) is the permeability of free space (\(4\pi \times 10^{-7} \, \text{T m/A}\)),
- \(N\) is the number of turns,
- \(I\) is the current,
- \(R\) is the radius of the coil.

### Step 2: Relate the magnetic field to the deflection angle
In a tangent galvanometer, the magnetic field \(B\) produced by the coil must balance the horizontal magnetic field \(B_H\) to achieve a deflection angle \(\theta\). The relationship is given by:
\[ B = B_H \tan(\theta) \]
For \(\theta = 60^\circ\), we have:
\[ \tan(60^\circ) = \sqrt{3} \]
Thus,
\[ B = B_H \sqrt{3} \]

### Step 3: Substitute \(B\) into the formula
Now, substituting for \(B\) in the first equation, we get:
\[ B_H \sqrt{3} = \frac{\mu_0 N I}{2R} \]

### Step 4: Solve for the current \(I\)
Rearranging the equation to solve for \(I\):
\[ I = \frac{2R B_H \sqrt{3}}{\mu_0 N} \]

### Step 5: Substitute the known values
Now, we can substitute the known values:
- \(R = 0.1 \, \text{m}\)
- \(N = 10\)
- \(B_H = 4 \times 10^{-5} \, \text{T}\)
- \(\mu_0 = 4\pi \times 10^{-7} \, \text{T m/A}\)

Substituting these values into the equation:
\[ I = \frac{2 \times 0.1 \times (4 \times 10^{-5}) \times \sqrt{3}}{4\pi \times 10^{-7} \times 10} \]

### Step 6: Simplify the expression
Calculating the numerator:
\[ 2 \times 0.1 \times (4 \times 10^{-5}) \times \sqrt{3} = 8 \times 10^{-6} \sqrt{3} \]
Calculating the denominator:
\[ 4\pi \times 10^{-7} \times 10 = 4\pi \times 10^{-6} \]
Thus,
\[ I = \frac{8 \times 10^{-6} \sqrt{3}}{4\pi \times 10^{-6}} = \frac{2\sqrt{3}}{\pi} \]

### Step 7: Calculate the value
Using \(\sqrt{3} \approx 1.732\) and \(\pi \approx 3.14\):
\[ I \approx \frac{2 \times 1.732}{3.14} \approx \frac{3.464}{3.14} \approx 1.1 \, \text{A} \]

### Final Answer
The current required to produce a deflection of \(60^\circ\) is approximately \(1.1 \, \text{A}\).

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