A cylindrical conductor of radius `'R'` carries a current `'i'`. The value of magnetic field at a point which is `R//4` distance inside from the surface is `10 T`. Find the value of magnetic field at point which is `4R` distance outside from the surface

Options

A. `4/3 T`
B. `8/3 T`
C. `40/3 T`
D. `80/3 T`

Answer

To solve the problem step by step, we will use the formulas for the magnetic field inside and outside a cylindrical conductor.

### Step 1: Understand the problem
We have a cylindrical conductor of radius \( R \) carrying a current \( i \). We need to find the magnetic field at a point that is \( 4R \) distance outside from the surface of the conductor, given that the magnetic field at a point \( \frac{R}{4} \) distance inside from the surface is \( 10 \, T \).

### Step 2: Define the distance for the inside point
The distance from the center of the cylinder to the point inside the conductor is:
\[ r = R - \frac{R}{4} = \frac{3R}{4} \]

### Step 3: Use the formula for the magnetic field inside the cylindrical conductor
The magnetic field \( B \) inside a cylindrical conductor is given by the formula:
\[ B_{\text{inside}} = \frac{\mu_0}{4\pi} \cdot \frac{2i r}{R^2} \]
Substituting \( r = \frac{3R}{4} \):
\[ B_{\text{inside}} = \frac{\mu_0}{4\pi} \cdot \frac{2i \cdot \frac{3R}{4}}{R^2} \]
This simplifies to:
\[ B_{\text{inside}} = \frac{\mu_0 \cdot 3i}{8\pi R} \]

### Step 4: Set the inside magnetic field equal to the given value
We know from the problem that:
\[ B_{\text{inside}} = 10 \, T \]
Thus, we can set up the equation:
\[ \frac{\mu_0 \cdot 3i}{8\pi R} = 10 \]

### Step 5: Use the formula for the magnetic field outside the cylindrical conductor
The magnetic field \( B \) outside a cylindrical conductor is given by:
\[ B_{\text{outside}} = \frac{\mu_0}{4\pi} \cdot \frac{2i}{r} \]
For a point \( 4R \) outside the surface, the distance \( r \) from the center of the cylinder is:
\[ r = R + 4R = 5R \]
Thus, we can write:
\[ B_{\text{outside}} = \frac{\mu_0}{4\pi} \cdot \frac{2i}{5R} \]

### Step 6: Relate the two magnetic fields
Now we can relate the two magnetic fields using the equations we derived:
\[ \frac{B_{\text{inside}}}{B_{\text{outside}}} = \frac{(R - \frac{R}{4})(R + 4R)}{R^2} \]
Substituting the values:
\[ \frac{10}{B_{\text{outside}}} = \frac{(\frac{3R}{4})(5R)}{R^2} \]
This simplifies to:
\[ \frac{10}{B_{\text{outside}}} = \frac{15R}{4R^2} = \frac{15}{4R} \]

### Step 7: Solve for \( B_{\text{outside}} \)
Rearranging gives:
\[ B_{\text{outside}} = \frac{10 \cdot 4R}{15} = \frac{40R}{15} = \frac{8R}{3} \]

### Step 8: Conclusion
Thus, the magnetic field at a point which is \( 4R \) distance outside from the surface of the cylindrical conductor is:
\[ B_{\text{outside}} = \frac{8}{3} \, T \]