Question

A current carrying coil is bent sharply so as to convert it into a double loop both carrying current in the same direction. If B is the initial magnetic field at the centre, then what will be the concentric final magnetic field
a) 2B
b) 4B
c) 8B
d) zero

Answer 1

Hint: It is given in the question that the coil is bent such that we form a double coil. This implies that the circumference of the cross section of the coil is half than that of the previous coil. Hence we will first determine the radius of the new resultant coil on bending and then substitute in the expression of magnetic field to determine the magnetic field due to the double loop in terms of the single loop.

Formula used:
$B=\dfrac{{{\mu }_{\circ }}i}{2R}T$
$C=2\pi R$
$B(\text{double loop})=B(coil1)+B(coil2)$

Complete answer:
Let us consider the coil of circular cross section with radius R and carrying current i with the permeability of free space ${{\mu }_{\circ }}$. Therefore the magnetic field at the centre of the coil is given by,
$B=\dfrac{{{\mu }_{\circ }}i}{2R}T$ . It is given in the question that the coil is bent into a double loop. Circumference of the bigger loop will be equal to twice the circumference of the single smaller double loop. Let the radius of the smaller loop be r. Therefore its circumference (C)in terms of the larger loop is written as,
$\begin{align}
  & C=2\pi R=2\left( 2\pi r \right) \\
 & \Rightarrow r=\dfrac{R}{2} \\
\end{align}$
Since the radius of the loop is R/2 and the coil contains two loops both the loops will generate magnetic fields. Which is given by,
$\begin{align}
  & B(\text{double loop})=B(coil1)+B(coil2) \\
 & \Rightarrow B(\text{double loop})=\dfrac{{{\mu }_{\circ }}i}{2\left( R/2 \right)}+\dfrac{{{\mu }_{\circ }}i}{2\left( R/2 \right)} \\
 & \Rightarrow B(\text{double loop})=2\dfrac{{{\mu }_{\circ }}i}{2R}+2\dfrac{{{\mu }_{\circ }}i}{2R}\text{, since }B=\dfrac{{{\mu }_{\circ }}i}{2R} \\
 & \Rightarrow B(\text{double loop})=2B+2B=4B \\
\end{align}$

Hence the correct answer of the above question is option b.

Note:
In the above case the coil consisted of a single circular loop. If the coil consists of n loops then the magnetic field by all the loops will be equal to $n\dfrac{{{\mu }_{\circ }}i}{2R}$. It is to be noted that all the loops should have their centre on the axis of the coil and the radius of each of the loops should be the same.