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An electron enters a uniform magnetic field of 0.5 T with a velocity of 2 ×10⁶ m/s perpendicular to the field. Calculate the radius of the circular path described by the electron. (Mass of electron = 9.1 ×10⁻³¹ kg, Charge = 1.6 ×10⁻¹⁹ C)
22.75 ×10⁻⁶ m
11.37 ×10⁻⁶ m
45.5 ×10⁻⁶ m
5.68 ×10⁻⁶ m
22.75 ×10⁻⁶ m
When a charged particle enters a uniform magnetic field perpendicular to its velocity, it experiences a Lorentz force perpendicular to both its motion and the magnetic field. This force acts as a centripetal force, causing the particle to move in a circular path with radius r=qBmv.
When a charged particle enters a uniform magnetic field perpendicular to its velocity, it experiences a Lorentz force perpendicular to both its motion and the magnetic field. This force acts as a centripetal force, causing the particle to move in a circular path with radius r=qBmv.
Think of a ball tied to a string being whirled in a circle; the tension in the string provides the inward pull necessary to keep the ball from flying off in a straight line, just as the magnetic force pulls the electron into a curve.
r=qBmv — Radius of circular orbit for a charged particle in a magnetic field
Fm=qvB — Magnitude of magnetic force (Lorentz force) when θ=90°
The magnetic Lorentz force is given by F=q(v×B). Since the velocity v is perpendicular to the magnetic field B, the magnitude of the force is F=qvB. Because this force is always perpendicular to velocity, it performs no work and changes only the direction of motion, acting as the centripetal force rmv2, leading to circular motion.
The magnetic force is always perpendicular to the velocity, so the speed of the electron remains constant.
The radius is directly proportional to momentum (p=mv) and inversely proportional to the magnetic field strength (B).
If the velocity were not perpendicular to the field, the path would be helical.
Used in particle accelerators like cyclotrons to confine and guide particles.
Cathode Ray Tubes (CRT) deflection systems
Mass Spectrometry for identifying ions based on their mass-to-charge ratio
Given data: m=9.1×10−31 kg, v=2×106 m/s, B=0.5 T, q=1.6×10−19 C.
Calculation: r=(1.6×10−19)×0.5(9.1×10−31)×(2×106)=0.8×10−1918.2×10−25=22.75×10−6 m.
Option B is calculated if one incorrectly uses q=3.2×10−19 or makes an error in powers of ten.
A is correct — The radius of the electron's circular path is calculated to be 22.75×10−6 m using the formula r=qBmv.
Always ensure your units are in SI (m, kg, s, C, T) before plugging values into the formula to avoid power-of-ten mistakes.