Question

how will magnetic field at centre of current carrying circular loop if current becomes double and radius of coil is halved

Answer 1

The magnetic field at a distance axially $x$ from the center of a current carrying loop of radius [tex] r

[/tex] is

$B(x)=\frac{\mu_0I}{2} \frac{r^2}{(x^2+r^2)^{3/2}}$.

The magnetic field at the center of the coil is

[tex] B(0)=\frac{\mu_0I}{2} \frac{r^2}{(0^2+r^2)^{3/2}}\\

B(0)=\frac{\mu_0I}{2} \frac{r^2}{r^3}\\

B(0)=\frac{\mu_0I}{2r} [/tex]

When $I'=2I,r'=\frac{r}{2}$

$B'(0)=\frac{\mu_02I}{2(r/2)}=4B(0)$

The magnetic field increases by a factor of 4.