Question

A current passing through a circular coil of two turns produces a magnetic field $B$ at its center. The coil then produces a magnetic field $B$ at its center. The coil is then rewound so as to have four turns and the same current is passed through it. The magnetic field at its center now is-
A) $2B$
B) $\dfrac{B}{2}$
C) $\dfrac{B}{4}$
D) $4B$

Answer 1

Hint:Magnetic field at the center of a coil is directly proportional to the number of turns it is having. If the current $(I)$ and radius $\left( R \right)$ of the coil is taken the same then the only variable is the number of turns $\left( n \right)$ on which magnetic field $\left( B \right)$ is depending.

Complete step by step solution:
If a coil of radius R is carrying current I then magnetic field produced at the center will be-
${\mathbf{B}} = \dfrac{{{\mu _0}nI}}{{2R}}$
In question it is given that 2 turns are producing magnetic field $B$
$ \Rightarrow $${\mathbf{B}} = \dfrac{{2{\mu _0}I}}{{2R}} = \dfrac{{{\mu _0}I}}{R}$
Hence it is magnetic field produced by two turns, so magnetic field produce by single turn will be half of $B$ i.e. ${B_1} = \dfrac{B}{2}$.
So, the magnetic field produced by 4 turns will be ${B_2} = 4{B_1} = 2B$.
In other words we can say that in the second case we are taking twice the number of turns than in the first case. So, the magnetic field produced will be twice as fast as in the first case.
$ \Rightarrow {B_2} = 2B$

Therefore, option A is the correct answer.

Note:In first case we are taking 2 turns to get magnetic field $B$ at the center of the coil and in second case it is 4 turns i.e. twice as in first case that’s why the output ${B_2}$ at the center is $2B$. If there is only one turn in first case then magnetic field produced at center with 4 turns will be 4 times i.e. $4B$ instead of $2B$.