A current carrying circular loop of radius 10 cm, has a magnetic induction of` 3.6 xx 10^(-5)T `at its center. What is the magnetic dilpoe moment of the loop?

Options

A. `150 xx 10^(-3) Am^(2)`
B. `180 xx 10^(-3) Am^(2)`
C. `160 xx 10^(-3) Am^(2)`
D. `145 xx 10^(-3) Am^(2)`

Answer

B

Explanation

`B=(mu_0)/(4pi)((2piI)/r)therefore3.6 xx 10^(-5)=10^(-7)x (2piI)/(10^(-1))therefore I=(18)/pitherefore `Magnetic dipole moment`IA=(18)/pi xx pir^(2)=18 xx 10^(-10)=180 xx 10^(-3)`.