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Inside a long solenoid carrying current $I$ with $n$ turns per unit length, the magnetic field is
A$\mu_0\,n\,I$
B$\mu_0\,I / n$
C$\mu_0\,n / I$
D$\mu_0\,n^2 I$
Answer & Solution
Correct answer: A. $\mu_0\,n\,I$
1. NCERT §4.5.2 derives the solenoid field from Ampere's law applied to a rectangular Amperian loop, partly inside and partly outside the solenoid.
2. Inside an IDEAL (infinitely long) solenoid, $\vec{B}$ is UNIFORM, directed along the axis, with magnitude $B = \mu_0\,n\,I$.
3. $n$ is the number of turns per unit length (not the TOTAL number of turns). If the solenoid has $N$ turns over length $L$, then $n = N/L$.
4. Outside the solenoid (ideal case), $B \approx 0$.
5. Real (finite-length) solenoids approach this value near their middle; field at the ends is half ($\mu_0 nI/2$).
6. Option B and C have wrong functional forms. Option D would give nonsense units.
_Source: NCERT Class 12 Physics Part 1, Ch 4, §4.5.2 (Magnetic Field of a Solenoid, Eq. 4.24), p. 16–17._
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