A coil carrying a current of `i=10mA` is placed in uniform magnetic field so that its axis consists of only one turn and is made of copper. The diameter of the wire is `0.1mm` , the radius of coil is `R=3cm`. An approximate external field B will rupture the coil is found `y xx10^(3)`. Find y, Breaking stress `=3 xx10^(8)N//m^(2)`.

Explanation

To solve the problem, we need to find the approximate external magnetic field \( B \) that will rupture the coil, given the parameters of the coil and the breaking stress of the material. Let's break down the solution step by step.

### Step 1: Understand the Given Data
- Current \( i = 10 \, \text{mA} = 10 \times 10^{-3} \, \text{A} \)
- Diameter of the wire \( d = 0.1 \, \text{mm} = 0.1 \times 10^{-3} \, \text{m} \)
- Radius of the coil \( R = 3 \, \text{cm} = 3 \times 10^{-2} \, \text{m} \)
- Breaking stress \( \sigma = 3 \times 10^{8} \, \text{N/m}^2 \)

### Step 2: Calculate the Cross-Sectional Area of the Wire
The radius of the wire \( r \) is half of the diameter:
\[ r = \frac{d}{2} = \frac{0.1 \times 10^{-3}}{2} = 0.05 \times 10^{-3} \, \text{m} \]
The cross-sectional area \( A \) of the wire is given by:
\[ A = \pi r^2 = \pi (0.05 \times 10^{-3})^2 = \pi (2.5 \times 10^{-9}) \approx 7.85 \times 10^{-9} \, \text{m}^2 \]

### Step 3: Calculate the Tension in the Wire
The breaking tension \( T \) can be calculated using the breaking stress:
\[ T = \sigma \cdot A = (3 \times 10^{8}) \cdot (7.85 \times 10^{-9}) \approx 2.355 \times 10^{0} \, \text{N} \approx 2.36 \, \text{N} \]

### Step 4: Relate Magnetic Force to Tension
The magnetic force \( F_m \) acting on the coil can be expressed as:
\[ F_m = i \cdot L \cdot B \]
where \( L \) is the length of the wire in the coil. For a single turn, the length \( L \) of the coil is the circumference:
\[ L = 2\pi R = 2\pi (3 \times 10^{-2}) \approx 0.1884 \, \text{m} \]

### Step 5: Set the Magnetic Force Equal to the Tension
At the breaking point, the magnetic force equals the tension:
\[ i \cdot L \cdot B = T \]
Substituting the values we have:
\[ (10 \times 10^{-3}) \cdot (0.1884) \cdot B = 2.36 \]

### Step 6: Solve for the Magnetic Field \( B \)
Rearranging the equation to solve for \( B \):
\[ B = \frac{2.36}{(10 \times 10^{-3}) \cdot (0.1884)} \approx \frac{2.36}{0.001884} \approx 124.8 \, \text{T} \]

### Step 7: Convert to the Required Format
The problem states that the answer should be in the form \( y \times 10^3 \):
\[ B \approx 124.8 \, \text{T} \approx 1.248 \times 10^2 \, \text{T} \approx 1.25 \times 10^3 \, \text{T} \]
Thus, \( y \approx 1.25 \).

### Final Answer
The value of \( y \) is approximately \( 1.25 \).

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