NDA (I) 2017 General Ability Test · Question 87 of 149

Correct — B, 2B. Inside a long current-carrying solenoid the magnetic field is uniform and is given by B = μ₀nI, where n is the number of turns per unit length and I the current. The field is therefore directly proportional to n with the current unchanged, so replacing n by 2n — twice as many turns packed into every centimetre of length — doubles the field to 2B.

• (a)B — The field would stay the same only if it did not depend on the turn density, but each turn contributes its own field and the contributions add along the axis, so packing more turns into the same length strengthens the field.
• (c)B/2 — Halving requires n in the denominator. Turn density multiplies the field, so more turns per unit length can never weaken it.
• (d)B/4 — A quarter would need an inverse-square dependence on n, which no solenoid formula contains. Even the field of a straight wire falls off only as one over the distance, and that is a dependence on distance, not on turn density.

A solenoid is a wire wound into a long helix. The circular fields of the individual turns reinforce one another along the axis and largely cancel outside, so the interior field is nearly uniform and parallel to the axis while the outside field is weak. Its magnitude is μ₀nI, which depends on how tightly the turns are packed and on the current, but not on the solenoid's radius and not on the distance from the axis inside it. A solenoid therefore behaves like a bar magnet with a north and a south end, and inserting a soft-iron core multiplies the field many times over, which is how an electromagnet is made.

The item is a one-line proportionality test, so read the formula and read the change. One printing point is worth noting: the paper writes 'If number of turn per unit length is increased to 2n', dropping the plural in the second sentence, but it is the same quantity n that the first sentence defined as the number of turns per unit length. The physical warning behind the formula is that n is turns per unit length, not the total number of turns — winding the same total number of turns over a longer former lowers n and weakens the field.

• The field inside a long solenoid is B = μ₀nI, with n the number of turns per unit length.
• That field is uniform along the interior and does not depend on the solenoid's radius.
• A current-carrying solenoid behaves like a bar magnet, with the field lines emerging from one end and entering the other.
• Placing a soft-iron core inside a solenoid greatly increases the field, which is the basis of the electromagnet.

• Reading n as the total number of turns instead of turns per unit length.
• Expecting the field to fall off away from the axis — inside a long solenoid it is essentially uniform.
• Confusing the solenoid formula with the field of a single circular loop, which does depend on the radius.

NDA asks either how B changes when n or I is scaled, or which statement about a current-carrying solenoid is not correct.

• practice — not a real PYQ

  The current through a long solenoid is doubled while the number of turns per unit length is unchanged. The magnetic field inside becomes

  • (a)unchanged
  • (b)half
  • (c)double
  • (d)one-fourth

  Answer(c) double — B = μ₀nI is directly proportional to the current as well as to the turn density.

• practice — not a real PYQ

  The magnetic field inside a long current-carrying solenoid is

  • (a)zero everywhere
  • (b)uniform and parallel to the axis
  • (c)strongest at the axis and zero at the walls
  • (d)directed radially outward

  Answer(b) uniform and parallel to the axis — the turns reinforce one another inside and largely cancel outside.