A solid cylinder carries current of 1 A uniformly distributed over its cross sectional area. Calculate the magnetic field induction at point P, at a distance of 3 cm from the axis of cylinder, if its radius is 7 cm .

Answer

To solve the problem of finding the magnetic field induction at point P, which is at a distance of 3 cm from the axis of a solid cylinder carrying a current of 1 A uniformly distributed over its cross-sectional area, we can follow these steps:

### Step 1: Identify the parameters
- Current (I) = 1 A
- Distance from the axis to point P (r) = 3 cm = 0.03 m
- Radius of the cylinder (R) = 7 cm = 0.07 m

### Step 2: Use Ampere's Circuital Law
According to Ampere's Circuital Law, the magnetic field (B) around a closed loop is given by:
\[ B \cdot DL = \mu_0 \cdot I_{\text{enclosed}} \]
Where:
- \( \mu_0 \) = permeability of free space = \( 4\pi \times 10^{-7} \, \text{T m/A} \)

### Step 3: Determine the enclosed current
Since the current is uniformly distributed over the cross-section of the cylinder, we need to find the current enclosed by a circular loop of radius 3 cm. The current density (J) can be calculated as:
\[ J = \frac{I}{A} = \frac{I}{\pi R^2} \]
Where \( A \) is the area of the cross-section of the cylinder.

The area of the entire cross-section of the cylinder is:
\[ A = \pi R^2 = \pi (0.07)^2 \]

Now, calculate the current density:
\[ J = \frac{1}{\pi (0.07)^2} \]

The area of the circular loop of radius 3 cm is:
\[ A_{\text{loop}} = \pi (0.03)^2 \]

The enclosed current \( I_{\text{enclosed}} \) in the loop is given by:
\[ I_{\text{enclosed}} = J \cdot A_{\text{loop}} \]

Substituting the values:
\[ I_{\text{enclosed}} = \left(\frac{1}{\pi (0.07)^2}\right) \cdot \pi (0.03)^2 \]

### Step 4: Simplify the expression for enclosed current
The \( \pi \) cancels out:
\[ I_{\text{enclosed}} = \frac{(0.03)^2}{(0.07)^2} \cdot 1 \]

Calculating this gives:
\[ I_{\text{enclosed}} = \frac{0.0009}{0.0049} \approx 0.1837 \, \text{A} \]

### Step 5: Apply Ampere's Law
Now we can substitute \( I_{\text{enclosed}} \) back into Ampere's Law:
\[ B \cdot (2\pi r) = \mu_0 \cdot I_{\text{enclosed}} \]

Where \( r = 0.03 \, \text{m} \):
\[ B \cdot (2\pi (0.03)) = (4\pi \times 10^{-7}) \cdot (0.1837) \]

### Step 6: Solve for B
Now, simplify:
\[ B \cdot (0.06\pi) = (4\pi \times 10^{-7}) \cdot (0.1837) \]

Dividing both sides by \( 0.06\pi \):
\[ B = \frac{(4 \times 10^{-7} \cdot 0.1837)}{0.06} \]

Calculating this gives:
\[ B \approx 1.224 \times 10^{-5} \, \text{T} \]

### Final Answer
The magnetic field induction at point P is approximately:
\[ B \approx 1.224 \times 10^{-5} \, \text{T} \]

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