Q. A length of wire carries a steady current. It is bent first to form a circular 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
A) Double of its first value
B) Quarter of its first value
C) Four times of its first value
D) Same as the first value
Explanation
(a) For first case, the wire of length $L$ is bent to form a circular coil of one turn,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ L = 2\pi r_{1}$
Similarly for second case,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ L = 4\pi r_{2}$
Now, $\ 2\pi r_{1} = 4\pi r_{2}\ or\ \ r_{2} = \frac{r_{1}}{2}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ B_{1} = \frac{\mu_{0}I}{2r_{1}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ B_{2} = \frac{\mu_{0}I}{2r_{2}} = \left( \frac{\mu_{0}I}{2r_{1}} \right) \times 2$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ B_{2} = 2B_{1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.