For a hollow metallic tube carrying current uniformly distributed over its surface,
(A) Magnetic field intensity is zero everywhere inside the tube.
(B) Magnetic field intensity inside the tube remains constant and is non-zero.
(C) Magnetic field intensity inside the tube is same as its value on surface.
(D) Magnetic field for points outside the tube is same as the magnetic field produced by a straight wire kept along the tube axis carrying same current.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (D) only
Apply Ampere’s circuital law, \( \oint\vec{B}\cdot d\vec{l} = \mu_{0}I_{\mathrm{enclosed}} \), on a circle of radius \( r \) coaxial with the tube of radius \( R \). By symmetry, \( B \) has the same value all round the circle.
\( r \lt R\text{: } B(2\pi r) = \mu_{0}\times0 \ \Rightarrow\ B = 0 \)
\( \displaystyle r \gt R\text{: } B(2\pi r) = \mu_{0}I \ \Rightarrow\ B = \frac{\mu_{0}I}{2\pi r} \)
(A) Magnetic field intensity is zero everywhere inside the tube. — Correct. Any circle inside the tube encloses no current, since all of it flows on the surface.
(B) Magnetic field intensity inside the tube remains constant and is non-zero. — Incorrect. The field inside is constant, but its value is zero.
(C) Magnetic field intensity inside the tube is same as its value on surface. — Incorrect. Inside it is zero. At the surface it is \( \mu_{0}I/(2\pi R) \), which is not zero.
(D) Magnetic field for points outside the tube is same as the magnetic field produced by a straight wire kept along the tube axis carrying same current. — Correct. Outside, the circle encloses the full current \( I \), so \( B = \mu_{0}I/(2\pi r) \), exactly the field of a long straight wire.