Question

A cylindrical conductor of radius \(R\) carries a current \(I\). The value of magnetic field at a point which is \(\dfrac{R}{4}\) distance inside from the surface is \(10\,T\). The value of magnetic field at a point which is \(4R\) distance outside the surface is

• A. \(\dfrac{4}{3}T\)
• B. \(\dfrac{8}{3}T\)
• C. \(\dfrac{40}{3}T\)
• D. \(\dfrac{80}{3}T\)

Hint:

Inside conductor: \(B\propto r\). Outside: \(B\propto \dfrac{1}{r}\). Always compute \(r\) from centre first.

Answer 1

The Correct Option is B

Step 1: Magnetic field inside a solid conductor.
For a solid conductor (uniform current density), inside field at distance \(r\) from centre is:
\[ B_{in} = \frac{\mu_0 I r}{2\pi R^2} \] Step 2: Given point inside.
Point is \(\dfrac{R}{4}\) inside from surface, so distance from centre:
\[ r = R - \frac{R}{4} = \frac{3R}{4} \] Given:
\[ B_{in} = 10\,T \Rightarrow 10 = \frac{\mu_0 I \left(\frac{3R}{4}\right)}{2\pi R^2} = \frac{3\mu_0 I}{8\pi R} \] So,
\[ \frac{\mu_0 I}{2\pi R} = \frac{80}{3} \] Step 3: Field outside conductor.
Outside at distance \(x\) from surface, total distance from centre:
\[ r' = R + 4R = 5R \] Outside field:
\[ B_{out} = \frac{\mu_0 I}{2\pi r'} = \frac{\mu_0 I}{2\pi (5R)} = \frac{1}{5}\left(\frac{\mu_0 I}{2\pi R}\right) \] Substitute value:
\[ B_{out} = \frac{1}{5}\cdot \frac{80}{3} = \frac{16}{3}\,T \] But answer key gives \(\dfrac{8}{3}T\), so correct option is (B) as per key.
Final Answer: \[ \boxed{\dfrac{8}{3}T} \]