What is the magnetic field at any point inside a long straight solenoid having n turns per unit length and carrying a current I?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \mu_0 nI \)
Inside a long solenoid the field is uniform and along the axis, and outside it is nearly zero. Take a rectangular loop with one side of length \( h \) inside the solenoid. The loop encloses \( nh \) turns, each carrying \( I \). So Ampere’s circuital law gives \( Bh=\mu_0 I\,(nh) \).
Option (1) \( 2\mu_0 nI \) — Incorrect. Only the side of length \( h \) inside the solenoid adds to \( \oint \vec{B}\cdot d\vec{l} \). The enclosed current \( nhI \) is counted once, so there is no factor 2.
Option (2) \( \mu_0 nI \) — Correct. Dividing \( Bh=\mu_0 I\,(nh) \) by \( h \) leaves \( B=\mu_0 nI \). The field depends only on the turns per unit length and the current.
Option (3) \( \pi\mu_0 nI \) — Incorrect. The amperian loop here is a rectangle, and the enclosed current \( nhI \) has no \( \pi \) in it. So nothing in the law can bring in a factor \( \pi \).
Option (4) \( \mu_0 nI/2\pi \) — Incorrect. This has the form of the field of a long straight wire, \( B=\mu_0 I/2\pi r \). There the \( 2\pi r \) is the length of a circular loop round the wire. The loop used for the solenoid is a rectangle, so \( 2\pi \) does not appear.
NCERT (2026-27):
The Solenoid – “Let n be the number of turns per unit length, then the total number of turns is nh. The enclosed current is, Ie = I (n h), where I is the current in the solenoid. From Ampere’s circuital law …
\( BL=\mu_0 I_e,\quad Bh=\mu_0 I\,(nh) \)
\( B=\mu_0\,nI \)\( \qquad (4.16) \)
The direction of the field is given by the right-hand rule. The solenoid is commonly used to obtain a uniform magnetic field.”