$A$ particle having a charge $50 \ e$ is revolving in a circular path of radius $0.4 \ m$ with $1 \ r.p.s.$ The magnetic field produced at the centre of the circle is $(\mu_0 = 4 \pi \times 10^{-7} \ SI \ units$ and $e = 1.6 \times 10^{-19} \ C)$.
- A$10^{-7} \mu_0$
- B$10^{-10} \mu_0$
- C$10^{-14} \mu_0$
- D$10^{-17} \mu_0$
Explore More
Similar Questions
In an atom, electrons revolve around the nucleus along a path of radius $0.72 \text{ Å}$, making $9.4 \times 10^{18}$ revolutions per second. The equivalent current is [given, $e = 1.6 \times 10^{-19} \text{ C}$]. (in $\text{A}$)
MediumKCET 2022
View SolutionTwo long straight conductors with currents $I_1$ and $I_2$ are placed along $X$ and $Y$-axes respectively. The equation of the locus of points of zero magnetic induction is:
Medium
View SolutionMagnetic field at a distance $r$ from an infinitely long straight conductor carrying a steady current varies as
Two mutually perpendicular insulated conducting wires carrying equal currents $I$ intersect at the origin. The resultant magnetic induction at point $P(2 \ m, 3 \ m)$ will be:
Medium
View Solution$A$ cell is connected between the points $A$ and $C$ of a circular conductor $ABCD$ of center $O$ with angle $AOC = 60^o$. If $B_1$ and $B_2$ are the magnitudes of the magnetic fields at $O$ due to the currents in $ABC$ and $ADC$ respectively,the ratio $B_1/B_2$ is
Difficult
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo