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

 

  • Option 1)

    \frac{\mu 0NI}{b}

     

     

     

  • Option 2)

    \frac{2\mu 0NI}{a}
     

  • Option 3)


     \frac{\mu 0NI\, In \left( \frac{b}{a} \right ) }{2\left ( b-a \right )}

  • Option 4)

    \frac{2\mu 0NI\, In \left( \frac{b}{a} \right ) }{\left ( b-a \right )}

 

Answers (1)

As we learnt in 

Magnetic Field due to Circular Current at the centre -

If a coil of radius r is carrying current i then magnetic feild at a distnace X from its centre is:

B_{centre}= \frac{\mu_{0} }{4\pi } \frac{2\pi Ni}{r} = \frac{\mu_{0} Ni}{2r}

- wherein

 

 Consider a small thickness dx at distance x from centre.

Number of turn in dx=\frac{N}{b-a}.dx

\Rightarrow dB=\frac{\mu_o dI}{2x}=\frac{\mu_o NI}{2(b-a)} .\frac{dx}{x}

 B=\frac{\mu_o NI}{2(b-a)} \int_{a}^{b}\frac{dx}{x}=\frac{\mu_o NI}{2(b-a)}ln(b/a)


Option 1)

\frac{\mu 0NI}{b}

 

 

 

incorrect

Option 2)

\frac{2\mu 0NI}{a}
 

incorrect

Option 3)


 \frac{\mu 0NI\, In \left( \frac{b}{a} \right ) }{2\left ( b-a \right )}

correct

Option 4)

\frac{2\mu 0NI\, In \left( \frac{b}{a} \right ) }{\left ( b-a \right )}

incorrect

Boost your Preparation for JEE Main 2021 with Personlized Coaching
 
Exams
Articles
Questions