Consider two thin identical conducting wires covered with very thin insulating material. One of the wires is bent into a loop and produces magnetic field $${B_1},$$ at its centre when a current $$I$$ passes through it. The ratio $${B_1}:{B_2}$$ is:
A.
$$1:1$$
B.
$$1:3$$
C.
$$1:9$$
D.
$$9:1$$
Answer :
$$1:3$$
Solution :
For loop $$B = \frac{{{\mu _0}nI}}{{2a}}$$
where, $$a$$ is the radius of loop.
Then, $${B_1} = \frac{{{\mu _0}I}}{{2a}}$$
Now, for coil $$B = \frac{{{\mu _0}I}}{{4\pi }}.\frac{{2nA}}{{{x^3}}}$$
at the centre $$x$$ = radius of loop
$$\eqalign{
& {B_2} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2 \times 3 \times \left( {\frac{I}{3}} \right) \times \pi {{\left( {\frac{a}{3}} \right)}^2}}}{{{{\left( {\frac{a}{3}} \right)}^3}}} = \frac{{{\mu _0}.3I}}{{2a}} \cr
& \therefore \frac{{{B_1}}}{{{B_2}}} = \frac{{\frac{{{\mu _0}I}}{{2a}}}}{{\frac{{{\mu _0}.3I}}{{2a}}}} \cr
& {B_1}:{B_2} = 1:3 \cr} $$
Releted MCQ Question on Electrostatics and Magnetism >> Magnetic Effect of Current
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