question_answer

A coil having N turns is wound tightly in the form of a spiral with inner and outer radii a and b respectively. When a current I passes through the coil, the magnetic field at the centre is                            [IIT-JEE (Screening) 2001]

A)            \[\frac{{{\mu }_{0}}NI}{b}\]  

B)            \[\frac{2{{\mu }_{0}}NI}{a}\]

C)            \[\frac{{{\mu }_{0}}NI}{2(b-a)}\ln \frac{b}{a}\]        

D)            \[\frac{{{\mu }_{0}}{{I}^{N}}}{2(b-a)}\ln \frac{b}{a}\]

Correct Answer: C

Solution :

                   Number of turns per unit width \[=\frac{N}{b-a}\] Consider an elemental ring of radius x and with thickness dx Number of turns in the ring \[=dN=\frac{Ndx}{b-a}\] Magnetic field at the centre due to the ring element \[dB=\frac{{{\mu }_{0}}(dN)i}{2x}=\frac{{{\mu }_{0}}i}{2}.\frac{Ndx}{(b-a)}.\frac{1}{x}\] \[\therefore \]Field at the centre           \[=\int_{{}}^{{}}{dB=\frac{{{\mu }_{0}}Ni}{2(b-a)}}\int_{a}^{b}{\frac{dx}{x}}\]           \[=\frac{{{\mu }_{0}}Ni}{2(b-a)}\ln \frac{b}{a}.\]