A proton of mass 1.67 ×10^-27 kg and charge 1.6 ×10^-19 C is projected with a speed of 2 ×10^6

step 1 Hello students in this question we have a proton which is having mass equals to 1 .67 into 10 to

A proton of mass 1.67 ×10^-27 kg and charge 1.6 ×10^-19 C...

A proton of mass $1.67 \times 10^{-27} \mathrm{~kg}$ and charge $1.6 \times 10^{-19} \mathrm{C}$ is projected with a speed of $2 \times 10^{6} \mathrm{~ms}^{-1}$ at an angle of $60^{\circ}$ to the $x$-axis. If a uniform magnetic field of $0.104 \mathrm{~T}$ is applied along $y$-axis, the path of proton is (a) a circle of radius $=0.2 \mathrm{~m}$ and time period $=2 \pi \times 10^{-7} \mathrm{~s}$ (b) a circle of radius $=0.1$ mand time period $=2 \pi \times 10^{-7}$ s (c) a helix of radius $0.1 \mathrm{~m}$ and time period $=2 \pi \times 10^{-7} \mathrm{~s}$ (d) a helix of radius $0.2 \mathrm{~m}$ and time period $=2 \pi \times 10^{-7} \mathrm{~s}$

A proton of mass $1.67 \times 10^{-27} \mathrm{~kg}$ and charge $1.6 \times 10^{-19} \mathrm{C}$ is projected with a speed of $2 \times 10^{6} \mathrm{~ms}^{-1}$
at an angle of $60^{\circ}$ to the $x$-axis. If a uniform magnetic field of $0.104 \mathrm{~T}$ is applied along $y$-axis, the path of proton is
(a) a circle of radius $=0.2 \mathrm{~m}$ and time period $=2 \pi \times 10^{-7} \mathrm{~s}$
(b) a circle of radius $=0.1$ mand time period $=2 \pi \times 10^{-7}$ s
(c) a helix of radius $0.1 \mathrm{~m}$ and time period $=2 \pi \times 10^{-7} \mathrm{~s}$
(d) a helix of radius $0.2 \mathrm{~m}$ and time period $=2 \pi \times 10^{-7} \mathrm{~s}$

Key Concepts

Radius of Curvature Calculation

The radius of the circular path traced by a charged particle in a magnetic field is derived by equating the magnetic force to the centripetal force necessary for circular motion. This radius, which depends on the mass, speed, charge of the particle, and the magnetic field strength, is key in predicting the path of the particle.

Helical Motion and Velocity Decomposition

If a charged particle enters a magnetic field with its velocity having both perpendicular and parallel components with respect to the field, the perpendicular component causes circular motion while the parallel component allows the particle to continue moving forward. The resulting path is a helix, reflecting the superposition of uniform motion along the field direction and circular motion perpendicular to it.

Period of Circular Motion

The time period of the circular motion of a charged particle in a magnetic field is given by the ratio of the particle’s mass to the product of its charge and the magnetic field strength. This period is independent of the particle's speed, highlighting a fundamental aspect of the dynamics of charged particles moving under the influence of magnetic fields.

Lorentz Force

The Lorentz force is the fundamental force acting on a charged particle when it moves through electric and magnetic fields. In the context of magnetic fields, the force is perpendicular to both the velocity of the particle and the magnetic field direction, which is crucial for understanding the deviation of the particle from a straight-line path.

Circular Motion in a Magnetic Field

When the component of a charged particle’s velocity is perpendicular to a uniform magnetic field, the particle experiences a constant magnetic force that acts as a centripetal force, causing it to move in a circular trajectory. This motion is characterized by a radius determined by balancing the magnetic force with the required centripetal force.

Transcript

Please give Ace some feedback