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?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $8\times 10^{-10} T$
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
$∵B=\frac{(4π×10^{-7})(0.01)}{2×0.5×5π}$
$=\frac{4π×10^{-9}}{5π}$
$=8×10^{-10} T$