Question

A length of wire carries a steady current. It is first bent to form a circular coil of one turn. The same length is now bent more sharply to give a loop of two turns of smaller radius. The magnetic field at the centre caused by the same current now will be:

• A. a quarter of its first value
• B. same as that of the first value
• C. four times the first value
• D. double of its first value

Answer 1

Answer: (C)

four times the first value

The magnetic field at the centre of the circular loop of radius r and turns N is given by,

$B=\frac{{\mu }_{0}IN}{2r}$

In first case, $B_1=\frac{{\mu }_{0}I}{2r_1}$

where $l=2\pi r_1$

In second case, $B_2=\frac{{\mu }_{0}I\times 2}{2r_2}$

where $l=2\pi r_2+2\pi r_2=4\pi r_2$

Now, $\frac{B_2}{B_1}=2\times \frac{r_1}{r_2}=2\times 2=4$ (as $r_1/r_2=2$)

So, $B_2=4B_1$