Question

A long copper wire is wound in the form of a coil of the radius $r$. A current of $2A$ is passed through this coil and the magnetic induction at the center of this coil is noted. The same wire is now folded end to end and a coil of the same radius r is prepared and the same current is passed through it. The magnetic induction at the center:
$(A)$ Will be doubled
$\left( B \right)$ Will be halved
$\left( C \right)$ Will remain the same
$\left( D \right)$ Will drop to zero.

Answer 1

Hint: In this question, A long wire is wound in the form of a coil of radius r and later the wire is now folded end to end and the coil of the radius r is prepared and the same current is passed through it.

Complete step by step answer:
A coil of wire designed to form a strong magnetic field inside the coil is called a solenoid. The magnetic field due to wires can become quite stronger by wrapping the wire repeatedly around a cylinder. The applied current and the number of turns per unit length both are directly proportional to the magnetic field within the solenoid.
Here, the strength of the magnetic field is diminished because the length of the solenoid is also decreased. That is, the magnetic induction at the center will be halved as the length of the solenoid and magnetic field strength are directly proportional.

Hence the right answer is in option $\left( B \right)$.

Note: 1. The number of turns N refers to the number of loops the solenoid has.
2. A stronger magnetic field can be produced by a greater amount of loops.
3. There will be no dependence on the diameter of the solenoid.
4. The field strength does not depend on the position inside the solenoid.