A long cylinder of wire of radius 'a' carries a current i distributed uniformly over its cross section. If the magnetic fields at distances `rlta and Rgta` from the axis have equal magnitude, then

Options

A. `a=(R+r)/2`
B. `a=sqrt(Rr)`
C. `a=Rr//(R+r)`
D. `a=R^2//r`

Explanation

To solve the problem, we need to find the relationship between the radius 'a' of a long cylindrical wire carrying a uniformly distributed current 'I', and the distances 'r' and 'R' from the axis of the cylinder, given that the magnetic fields at these distances have equal magnitudes.

### Step-by-Step Solution:

1. **Understanding the Magnetic Field Inside the Cylinder (r < a)**:
- For a point inside the cylinder (where \( r < a \)), the magnetic field \( B_r \) can be calculated using Ampere's Law:
\[ B_r \cdot 2\pi r = \mu_0 I_{\text{enclosed}} \]
- The enclosed current \( I_{\text{enclosed}} \) for a radius \( r \) is given by:
\[ I_{\text{enclosed}} = I \cdot \frac{\pi r^2}{\pi a^2} = \frac{I r^2}{a^2} \]
- Substituting this into Ampere's Law gives:
\[ B_r \cdot 2\pi r = \mu_0 \cdot \frac{I r^2}{a^2} \]
- Therefore, the magnetic field \( B_r \) is:
\[ B_r = \frac{\mu_0 I r}{2\pi a^2} \]

2. **Understanding the Magnetic Field Outside the Cylinder (r > a)**:
- For a point outside the cylinder (where \( r > a \)), the magnetic field \( B_R \) is given by:
\[ B_R \cdot 2\pi R = \mu_0 I \]
- Thus, the magnetic field \( B_R \) is:
\[ B_R = \frac{\mu_0 I}{2\pi R} \]

3. **Setting the Magnetic Fields Equal**:
- According to the problem, the magnitudes of the magnetic fields at distances \( r \) and \( R \) are equal:
\[ B_r = B_R \]
- Substituting the expressions we derived:
\[ \frac{\mu_0 I r}{2\pi a^2} = \frac{\mu_0 I}{2\pi R} \]
- We can cancel \( \mu_0 I \) and \( 2\pi \) from both sides:
\[ \frac{r}{a^2} = \frac{1}{R} \]

4. **Rearranging the Equation**:
- Rearranging the equation gives:
\[ r = \frac{a^2}{R} \]

5. **Finding the Relationship Between a, r, and R**:
- From the equation \( r = \frac{a^2}{R} \), we can express \( a \) in terms of \( r \) and \( R \):
\[ a^2 = rR \]
- Taking the square root gives:
\[ a = \sqrt{rR} \]

### Final Answer:
Thus, the relationship between the radius 'a' of the wire, and the distances 'r' and 'R' from the axis is:
\[ a = \sqrt{rR} \]