Question

When current is passed through a circular wire prepared from a long conducting wire, the magnetic field produced at its centre is $B$. Now a loop having two turns is prepared from the same wire and the same current is passed through it. The magnetic field at its centre will be

• A. $4B$
• B. $\frac{B}{4}$
• C. $\frac{B}{2}$
• D. $16B$

Answer 1

Answer: (A)

$4B$

Magnetic field at the centre of current carrying coil is given by $B=\frac{{\mu }_0}{4\pi }\frac{2\pi Ni}{r}$
$\Rightarrow B\propto \frac{N}{r}$
$\therefore \frac{B_1}{B_2}=\frac{N_1}{N_2}\times \frac{r_2}{r_1}$

In second condition, a loop having two turns is prepared from the same wire
So, $N_1=N$
$N_2=2N$
$r_1=r$
$r_2=2r$
$\Rightarrow \frac{B}{B_2}=\frac{1}{2}\times \frac{r/2}{r}=\frac{1}{4}$
$\Rightarrow B_2=4B$.