Question

A current $i$ is flowing in a specific wire. It is turned into a circular coil of one turn. Then it is turned to make a coil of two turns and smaller radius. Now the magnetic induction at the centre for same current will be

• A. half of its previous value
• B. one fourth of its previous value
• C. four times of its previous value
• D. zero

Answer 1

Answer: (C)

four times of its previous value

Let the radii be $r_1$ and $r_2$ respectively.
Since there are two turns of radius $r_2$, $r_1=2r_2$
Magnetic field $B$ at the center of the coil of radius $r_1$ $B_1=\frac{{\mu }_{o}i}{2r_1}=\frac{{\mu }_{o}i}{4r_2}$
Magnetic field $B$ at the center of the coil of radius $r_2$ $B_2=2\times \frac{{\mu }_{o}i}{2r_2}$
$\therefore \frac{B_2}{B_1}=\frac{2\times \frac{{\mu }_{o}i}{2r_2}}{\frac{{\mu }_{o}i}{4r_2}}=4$