A charged particle with charge 0.01 C rotates on a circular path of radius 0.5 m, with velocity 0.2 m/s. What is the value of B at the center of the circle?

$8 \pi \times 10^{-10} T$

The correct answer is Option (1) → $8 \pi \times 10^{-10} T$

The equivalent current (I) is the charge passing (q) through the 100 per unit time (t).

$I=\frac{q}{t}$  [q = 0.01 C [given]]

Also,

$t=\frac{circumference}{velocity}=\frac{2π(0.5)}{0.2}=\frac{π}{0.2}$

$t=5π\,s$

$∴I=\frac{0.01}{5π}A$

Now,

$B=\frac{μ_0I}{2r}$

where, B = Magnetic field

$∵B=\frac{(4π×10^{-7})(0.01)}{2×0.5×5π}$

$=\frac{4π×10^{-9}}{5π}$

$=8×10^{-10} T$