Question

A coil of n number of turns is wound tightly in the form of a spiral with inner and outer radii a and b respectively. When a current of strength I is passed through the coil, the magnetic field at its centre is

• A. ${\mu }_{0}nI(b-a){log}_e\frac{a}{b}$
• B. $\frac{{\mu }_{0}nI}{2ab}$
• C. $\frac{2{\mu }_{0}nI}{b}$
• D. $\frac{{\mu }_{0}nI}{2(b-a)}$ ${log}_e\frac{b}{a}$

Answer 1

The correct option is D $\frac{{\mu }_{0}nI}{2(b-a)}$ ${log}_e\frac{b}{a}$

Magnetic field due to a coil of radius $r$ at its center $=\frac{{\mu }_{o}I}{2r}$

The thickness of the wire$\left(t\right)=\frac{b-a}{n}$

Let us choose a loop at radius $x$ and thickness $dx$

The thickness through $dx$ is $I\frac{dx}{T}$

Using this in formula:

We get magnetic field due to $dx$ is $\frac{{\mu }_{o}Idx}{2xt}$

For total magnetic field we need to integrate from $a$ to $b$ and substitute value of $t$.

Net magnetic field $={\int }_a^b\frac{{\mu }_{o}Idx}{2xt}$

Magnetic Field $=\frac{{\mu }_{o}nI{\log }_e\left(\frac{b}{a}\right)}{2(b-a)}$