Question

A wire of length $l$ carries a steady current. It is bent first to form a circular plane loop of one turn. The magnetic field at the centre of the loop is $B$. 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 is

• A. B
• B. B/4
• C. 4B
• D. B/2

Answer 1

Answer: (C)

4B

Answer 2

The wire of length $l$ is bent to from a circular loop, so $2\pi r = l$ $\Rightarrow\,r = \frac{l}{2\pi}$ The magnetic field at the centre of the loop is $B = \frac{\mu_{0}I}{2r} = \frac{\mu_{0}I\times2\pi}{2l}$ Now the same length of the wire is bent to from a double loop $\therefore 2\times2\pi\, r' = l$ $\Rightarrow\,r' = \frac{l}{4\pi}$ And the magnetic field at the centre $B' = \frac{\mu_{0}I\times2}{2\times r'}$ $= \frac{2\mu_{0}I}{2\times\frac{l}{4\pi}} = \frac{2\times4\pi\mu I}{2l}$ $\therefore\, \frac{B'}{B} = 4 \,\Rightarrow \,B' = 4B$