A certain particle is sent into a uniform magnetic field,...

A certain particle is sent into a uniform magnetic field, with the particle's velocity vector perpendicular to the direction of the field. Figure $28-37$ gives the period $T$ of the particle's motion versus the inverse of the field magnitude $B$ . The vertical axis scale is sct by $T_{s}=40.0 \mathrm{ns},$ and the horizontal axis scale is sct by $B_{s}^{-1}=5.0 \mathrm{T}^{-1} .$ What is the ratio $\mathrm{m} / \mathrm{q}$ of the particle's mass to the magnitude of its charge?

Key Concepts

Graphical Analysis of Linear Relationships

Plotting the period T of the particle's motion against the inverse of the magnetic field (1/B) results in a linear relationship. In this graph, the slope is equal to 2?m/q. Analyzing the graph allows one to extract the ratio of the particle's mass to its charge by simply measuring the slope and using the known relationship.

Cyclotron Frequency and Period

In cyclotron motion, the particle's angular frequency (or cyclotron frequency) is given by ? = qB/m, and the corresponding period of motion is T = 2?m/(qB). This relation is crucial because it directly ties the measurable period of the circular motion to the intrinsic properties of the particle, namely its mass and charge.

Lorentz Force

This concept refers to the force experienced by a moving charged particle in a magnetic field. The force is given by F = q(v × B), meaning that its magnitude is proportional to the charge, the velocity of the particle, and the strength of the magnetic field, while its direction is perpendicular to both the velocity and the field. This force is fundamental in determining the particle’s motion in a magnetic environment.

Circular Motion in a Uniform Magnetic Field

When a charged particle enters a uniform magnetic field with a velocity perpendicular to the field, the Lorentz force acts as a centripetal force, causing the particle to move in a circular path. The balance between the magnetic force and the centripetal force is what dictates the radius of the path and the properties of its motion.

Transcript

00:01 Hello and welcome to this video solution of numerate.

00:03 Here it's given that a certain particle is sent into a uniform magnetic field with the particle's velocity vector perpendicular to the direction of the field.

00:12 Figure 20 -37 gives you the period time of the particle versus the inverse of the magnetic field b.

00:22 So what it's given is, if you see this graph, then you have got t s along the y -axis and inverse of the magnetic field along the y -axis.

00:32 And it's clearly a straight line.

00:40 The time is in nanosecond and b is in tesla.

00:43 This is in nanoseconds and this is in tesla inverse.

00:47 Now based on this, the vertical axis is set by t s which is 4 .0 nanoseconds.

00:59 The scale is given and the horizontal scale is b s inverse is 5 .0 tesla inverse.

01:11 Now you have to calculate the ratio of m by q of the particle to the mass by the charge.

01:18 So let's see how much we're getting.

01:21 So what you can do is, you can use the expression of time period t of what from where.

01:36 So what you can do is you can actually apply, if you remember, there is a motion of charge inside a magnetic field.

01:47 So here what happens is the charge involves circular motion which is which actually is due to the centripetal force m v square by r.

01:57 Right.

01:58 Or maybe what i can do is m omega square r i take.

02:03 Right.

02:03 That is equal to q v b.

02:09 Right.

02:13 This is a magnetic force.

02:14 Right.

02:15 And here you will be having m times of 2 pi by t whole squared times r will be equal to q times of v is r omega again.

02:28 Right.

02:29 So you can cancel rr and 1 omega like this times b.

02:35 Right.

02:36 And finally, you will be having the expression m by q that is equal to 2 pi.

02:50 Sorry m q will be equal to b by 2 pi t.

03:01 Right.

03:02 Sorry 2 pi on the upper side you have got t.

03:05 Right.

03:08 T b by 2 pi.

03:09 Now simply what you can do is this is equal to t by 2 pi b inverse.

03:20 Right.

03:22 Now from here you have to find the slope...

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