Question

The magnetic field of a given length of wire carrying a current for a single turn cicular coil at center is B, then its value for two turns for the same wire when same current passing through it is

• A. $\frac{B}{4}$
• B. $\frac{B}{2}$
• C. $2B$
• D. $4B$

Answer 1

The correct option is D $4B$

The magnetic field at the center of the coil is given by $B=\frac{\mu nI}{2a}$

where a and n is radius and no. of turn respectively

Total lenght of a coil is L

in 1 turn the radius is $2\pi a=L$

$\Rightarrow a=\frac{L}{2\pi }$

if no. of turn become twice than

$\Rightarrow 4\pi r=L$

$\Rightarrow r=\frac{L}{4\pi }=\frac{a}{2}$

n=2,radius become $\frac{a}{2}$

$\Rightarrow B_1=\frac{u2I}{\frac{a}{2}}=\frac{4\mu nI}{a}=4B$

hence the magnetic field become 4B