Question
The magnetic induction at a point $$P$$ which is at the distance of $$4\,cm$$ from a long current carrying wire is $${10^{ - 3}}T.$$ The field of induction at a distance $$12\,cm$$ from the current will be
A.
$$3.33 \times {10^{ - 4}}T$$
B.
$$1.11 \times {10^{ - 4}}T$$
C.
$$3 \times {10^{ - 3}}T$$
D.
$$9 \times {10^{ - 3}}T$$
Answer :
$$3.33 \times {10^{ - 4}}T$$
Solution :
Magnetic field due to a long straight conductor carrying current $$i$$ at a distance $$r$$ is given by
$$B = \frac{{{\mu _0}}}{{4\pi }}\frac{{2i}}{r}$$
Given, $${r_1} = 4\,cm,\,{r_2} = 12\,cm.$$
As $$B \propto \frac{1}{r}$$
and distance becomes 3 times, field is reduced to its one-third value.
Hence, $$B' = \frac{B}{3} = \frac{{{{10}^{ - 3}}}}{3} = 3.33 \times {10^{ - 4}}T$$