A length $L$ of wire carries a steady current $I$. It is bent first to form a circular plane coil of one turn. The same length is now bent more sharply to give a double loop of smaller radius. The magnetic field at the centre caused by the same current is
AIIMS 1980, Medium
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(c) Magnetic field at the centre of current carrying coil is given by $B = \frac{{{\mu _0}}}{{4\pi }} \cdot \frac{{2\pi Ni}}{r}$ $==>$ $B \propto \frac{N}{r}$ $==>$ $\frac{{{B_1}}}{{{B_2}}} = \frac{{{N_1}}}{{{N_2}}} \times \frac{{{r_2}}}{{{r_1}}}$.
The following figure shows that single turn coil changes to double turn coil.
$N_1 = 1 \,\,\,N_2 = 2$
$r_1 = r \,\,\,r_2 = r / 2$
$B_1 = B \,\,\,B_2 = ?$
$==>$ $\frac{B}{{{B_2}}} = \frac{1}{2} \times \frac{{r/2}}{r} = \frac{1}{4}$ $==>$ ${B_2} = 4B$
Short trick : For such type of problems remember ${B_2} = {n^2}{B_1}$
The following figure shows that single turn coil changes to double turn coil.
$N_1 = 1 \,\,\,N_2 = 2$
$r_1 = r \,\,\,r_2 = r / 2$
$B_1 = B \,\,\,B_2 = ?$
$==>$ $\frac{B}{{{B_2}}} = \frac{1}{2} \times \frac{{r/2}}{r} = \frac{1}{4}$ $==>$ ${B_2} = 4B$
Short trick : For such type of problems remember ${B_2} = {n^2}{B_1}$
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