In a solenoid, the current flowing through the wire is I and number of turns per unit length is n. This gives a magnetic field B inside the solenoid. If number of turns per unit length is increased to 2n, what will be the value of magnetic field in the solenoid ?

The correct answer is B, 2B. The field inside a long solenoid is directly proportional to the number of turns per unit length, so doubling n doubles the field.

For a long solenoid the magnetic field inside is B = μ₀nI, where μ₀ is the permeability of free space, n the turns per unit length and I the current. The field is nearly uniform inside and almost zero outside, which is why a solenoid behaves like a bar magnet. With the current unchanged and n raised to 2n, the new field is μ₀(2n)I = 2B. Putting a soft iron core inside raises the field far more, which is how an electromagnet is made.

A is wrong because the field does depend on the number of turns. C is wrong because it treats the field as inversely proportional to n. D is wrong because nothing in the formula falls with the square of n.

Exam tip: solenoid field B = μ₀nI; it doubles if either the turns per unit length or the current is doubled.