A proton is projected with a speed of 2 ×10^6( m / s) at an angle of 60^∘ to the X -axis. If a u

A proton is projected with a speed of 2 ×10^6( m / s) at an...

A proton is projected with a speed of $2 \times 10^{6}(\mathrm{~m} / \mathrm{s})$ at an angle of $60^{\circ}$ to the $\mathrm{X}$ -axis. If a uniform mag. field of $0.104$ tesla is applied along $\mathrm{Y}$ -axis, the path of proton is (a) A circle of $\mathrm{r}=0.2 \mathrm{~m}$ and time period $\pi \times 10^{-7} \mathrm{sec}$ (b) A circle of $\mathrm{r}=0.1 \mathrm{~m}$ and time period $2 \pi \times 10^{-7} \mathrm{sec}$ (c) A helix of $r=0.1 \mathrm{~m}$ and time period $2 \pi \times 10^{-7} \mathrm{sec}$ (d) A helix of $\mathrm{r}=0.2 \mathrm{~m}$ and time period $4 \pi \times 10^{-7} \mathrm{sec}$

A proton is projected with a speed of $2 \times 10^{6}(\mathrm{~m} / \mathrm{s})$ at an angle of $60^{\circ}$ to the $\mathrm{X}$ -axis. If a uniform mag. field of $0.104$ tesla is applied along $\mathrm{Y}$ -axis, the path of proton is
(a) A circle of $\mathrm{r}=0.2 \mathrm{~m}$ and time period $\pi \times 10^{-7} \mathrm{sec}$
(b) A circle of $\mathrm{r}=0.1 \mathrm{~m}$ and time period $2 \pi \times 10^{-7} \mathrm{sec}$
(c) A helix of $r=0.1 \mathrm{~m}$ and time period $2 \pi \times 10^{-7} \mathrm{sec}$
(d) A helix of $\mathrm{r}=0.2 \mathrm{~m}$ and time period $4 \pi \times 10^{-7} \mathrm{sec}$

Key Concepts

Time Period and Radius Formulas

The time period of the circular component of the motion is given by T = (2?m)/(qB) and the radius by r = (mv)/(qB), where m is the mass, q is the charge, v is the speed (or its perpendicular component), and B is the magnetic field strength. These formulas are key in determining the characteristics of the circular or helical paths.

Helical Motion

When a charged particle has both perpendicular and parallel velocity components relative to a magnetic field, it follows a helical trajectory. The circular motion determined by the perpendicular component combines with uniform linear motion along the field direction due to the unaffected parallel component.

Lorentz Force

This is the force experienced by a charged particle moving in a magnetic field, calculated by the cross product of the particle's velocity and the magnetic field. It is fundamental in determining the motion of the particle and is given by the expression F = q(v × B).

Decomposition of Velocity Components

When a charged particle is projected at an angle relative to the magnetic field, its velocity can be split into components perpendicular and parallel to the field. The perpendicular component is responsible for circular motion, while the parallel component remains unaffected by the magnetic field, leading to a combined helical motion.

Circular Motion in Magnetic Fields

The perpendicular component of a charged particle’s velocity in a magnetic field causes it to move in a circular path due to the constant magnitude of the Lorentz force acting as a centripetal force. The relationship among the radius, speed, magnetic field strength, and particle charge is critical for analyzing this motion.

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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