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

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 $2 \times 10^{6} \mathrm{~m} / \mathrm{s}$ 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 $\pi \times 10^{-7} \mathrm{~s}$. (B) a circle of radius $=0.1 \mathrm{~m}$ and time period $2 \pi \times 10^{-7} \mathrm{~s}$. (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 $4 \pi \times 10^{-7} \mathrm{~s}$.

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 $2 \times 10^{6} \mathrm{~m} / \mathrm{s}$ 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 $\pi \times 10^{-7} \mathrm{~s}$.
(B) a circle of radius $=0.1 \mathrm{~m}$ and time period $2 \pi \times 10^{-7} \mathrm{~s}$.
(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 $4 \pi \times 10^{-7} \mathrm{~s}$.

Key Concepts

Lorentz Force

The Lorentz force is the force experienced by a charged particle moving in a magnetic field and is given by F = q(v × B), where q is the charge, v is the velocity, and B is the magnetic field. This force acts perpendicular to both the velocity and the magnetic field, influencing the particle's path without doing any work on it, since it does not change the particle's speed.

Velocity Decomposition

When a charged particle enters a magnetic field, its velocity can be decomposed into two components: one parallel to the magnetic field and one perpendicular to it. The parallel component remains unaffected by the magnetic field, while the perpendicular component experiences the Lorentz force, resulting in circular motion.

Cyclotron Motion

Cyclotron motion refers to the circular motion of the charged particle due to the Lorentz force acting on the perpendicular component of its velocity in a uniform magnetic field. The radius of the circular path is given by r = m*v_perp/(qB) and the time period of the motion is T = 2?m/(qB), where m is the mass, v_perp is the perpendicular velocity component, q is the charge, and B is the magnetic field.

Helical Motion

Helical motion occurs when a charged particle has a velocity component both perpendicular and parallel to a magnetic field. While the perpendicular component causes circular motion, the parallel component leads to a constant linear motion along the direction of the magnetic field, resulting in a helical trajectory.

Transcript

00:01 Okay, as per the question, a proton of mass 1 .67 multiplied by 10 -rest -to -the -power minus 27 kg, that is given and with having the charge q, that is 1 .6 multiplied by 10 -rest -to -the -power minus 19 kulum.

00:15 Projected with a speed, v is given to multiplied by 10 -rest -to -the -power 6 meter per second at an angle of 60 degree to the x -axis.

00:23 So, theta is given, okay? 60 degree we can say theta is given 60 degree with the x -axis okay with the x -axis and if a uniform magnetic field of 0 .104 tesla so b is given 0 .104 tesla is applied along the y -axis okay so let's say this is the y -axis and this is the x -axis okay of course these are perpendicular and the magnetic field in the direction of y axis okay and the the particle or we can say the proton is projected 60 degree to the axis so it is like this okay it is 60 degree to the axis so here it is v so we will have the two components that in vx and v y okay if this is v then this will be vx and it will be y okay so vx it is 60 degree so it will be 30 degree okay so here it will be v sign 30 okay and here it will be v cost 30 okay so due to this vx the the path will be circle okay but when this the y component is there that will goes like this and it will become helical path okay so first of all the path will be helic due to the vx and vy both components are present there okay and now we know that how can we find out the radius of that helical path okay that is we know that it will be the radius we know that it is mv divided by qb okay but v that will be the perpendicular to the b okay that is this okay that is vx component or we can say it will be v sine 30 okay okay so here i am writing down v -sign theta where theta is 30 degree.

02:31 I'm writing down 30 degree here.

02:33 So we can calculate the radius here.

02:36 Now m that is given 1 .67 multiplied by 10x to the power minus 27.

02:41 V that is the velocity or speed it is given to multiplied by 10x to the power 6 meter per second...

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