motion in a uniform magnetic field continued force on a...

21/08/2013

Today’s lecture:

PHYSICS 1B

Electricity & Magnetism

Motion in a Uniform Magnetic Field continued

Force on a Current Carrying Conductor

Introduction to the Biot-Savart Law

Consider a particle moving in an external magnetic field with its velocity perpendicular to the field. The force the particle experiences is always directed towards the centre of the circular path. In other words: The magnetic force causes a centripetal acceleration, which changes the velocity of the particle. Remember: the speed of the particle doesn’t change.

21/08/2013

A Charged Particle in a Magnetic Field

PHYSICS 1B – Magnetism

Electricity & Magnetism – Magnetism

Use the particle under a net force and particle in uniform circular motion models (from mechanics) Equating the magnetic and centripetal forces:

𝐹𝐵 = 𝑞𝑣𝐵 =𝑚𝑣2

𝑟

Solve for r:

𝑟 =𝑚𝑣

𝑞𝐵

r is proportional to the particle’s linear momentum r is inversely proportional to the magnetic field

21/08/2013

Force on a Charged Particle



We can go further than this. The angular speed of the particle is clearly:

𝜔 =𝑣

𝑟=𝑞𝐵

𝑚

ω is often called the cyclotron frequency. The period of the motion is therefore:

𝑇 =2𝜋𝑟

𝑣=2𝜋

𝜔=2𝜋𝑚

𝑞𝐵

21/08/2013

A Charged Particle’s Motion



If our charged particle is moving in a magnetic field at some arbitrary angle with respect to the field, then the path it follows will be a helix. For the case where the field is in the +x direction: The same equations as before apply, but with the velocity v replaced by This gives the motion of the particle around the centre of the helix – its velocity in the x-direction does not change.

21/08/2013

A Charged Particle’s Motion: General



2 2

y zv v v

Electrons are accelerated from rest through a potential difference. These electrons then enter a uniform magnetic field that is perpendicular to their velocity vector. The electrons then travel in a curved path. We can use conservation of energy to find v, and from this we can the work out other parameters. See example 29.3 in the text book if you want to try working this one out!

21/08/2013

A Charged Particle’s Motion: Examples

The bending of an Electron Beam



The Van Allen radiation belts consist of charged particles surrounding the Earth in doughnut-shaped regions. Existence confirmed by James Van Allen, who was working on the Explorer 1 and 3 missions. The particles are trapped by the Earth’s non- uniform magnetic field. The particles spiral from pole to pole, and can create aurorae when they hit the Earth’s atmosphere.

21/08/2013

A Charged Particle’s Motion: Examples

The Van Allen Radiation Belts



21/08/2013

The Earth’s Magnetosphere and the Van Allen belts



In many applications, charged particles will move in the presence of both magnetic and electric fields. In such cases, the total force experienced by the particle is the sum of the forces due to the individual fields. The total force experienced by the particle due to the magnetic and electric fields is known as the Lorentz force. In general:

𝑭 = 𝑞𝑬 + 𝑞𝒗 × 𝑩

21/08/2013

Charged Particles Moving in E and B Fields.



A velocity selector is a device used in experiments which require charged particles that are all moving at the same velocity. A uniform electric field is perpendicular to a uniform magnetic field. When the force due to the electric field is equal in magnitude, but opposite in direction to the force due to the magnetic field, the particles will move in a straight line.

21/08/2013

The Velocity Selector



Particles will move in a straight line when:

𝑣 =𝐸

𝐵

Only the particles with the chosen speed will pass through undeflected. The fastest particles will deflect to the left (since for them, FB > FE

Slow moving particles will deflect to the right.

21/08/2013

The Velocity Selector



A mass spectrometer separates ions according to their mass-to-charge ratio. In the design to the right, a beam of ions passes through a velocity selector (so all have the same velocity). Those ions then enter a second magnetic field. In that field, the ions move in a semicircle of radius r before they strike a detector at point P.

21/08/2013

The Mass Spectrometer



Positively charged ions deflect to the left. Negatively charged ions deflect to the right. Check this with the right hand rule! The mass to charge ratio (m/q) can be determined if we know E and B, just by measuring r.

𝑚

𝑞=𝑟𝐵0𝑣=𝑟𝐵0𝐵

𝐸

21/08/2013

The Mass Spectrometer



J. J. Thomson measured the charge-mass ratio of the electron in 1897, using the apparatus to the right. Electrons are accelerated from the cathode. They are then deflected by electric and magnetic fields. The beam of electrons strikes a fluorescent screen. By measuring the beam’s deflection, it was possible to calculate e/m!

21/08/2013

Measuring the charge-mass ratio of the electron



A charged particle is moving perpendicular to a magnetic field in a circle, with radius r. An identical particle enters the field, again with v perpendicular to B, but with a greater speed, v, than the first particle. Compared to the radius of the circle followed by the first particle, the radius of the circle for the second particle is: a) Smaller

b) Larger

c) Equal in size

21/08/2013

Quick Quiz



A charged particle is moving perpendicular to a magnetic field in a circle, with radius r. An identical particle enters the field, again with v perpendicular to B, but with a greater speed, v, than the first particle. Compared to the radius of the circle followed by the first particle, the radius of the circle for the second particle is:

b) Larger The magnetic force on the particle increases in proportion to v, but the centripetal acceleration increases according to the square of v. Taken together, this results in a larger radius, as we can see from:

𝑟 =𝑚𝑣

𝑞𝐵

21/08/2013

Quick Answer



A charged particle is moving perpendicular to a magnetic field in a circle, with radius r. The magnitude of the magnetic field is increased. Compared to the initial radius of the circular path, the radius of the new path is: a) Smaller

b) Larger

c) Equal in size 21/08/2013

Quick Quiz



A charged particle is moving perpendicular to a magnetic field in a circle, with radius r. The magnitude of the magnetic field is increased. Compared to the initial radius of the circular path, the radius of the new path is: a) Smaller

The magnetic force on the particle increases in proportion to B. This results in a smaller radius for the circle followed by the particle. Again:

𝑟 =𝑚𝑣

𝑞𝐵

21/08/2013

Quick Answer



Three types of particle enter a mass spectrometer like that to the right. The figure below shows where the particles hit the detector. Rank the particles that arrive at a, b and c by speed. a) a, b, c b) b, c, a c) c, b, a d) All the speeds are equal

21/08/2013

Quick Quiz



Rank the particles that arrive at a, b and c by speed. d) All the speeds are equal The velocity selector ensures that all three types of particle have the same speed.

21/08/2013

Quick Answer



Now, rank the particles that arrive at a, b and c by their mass to charge ratio, m/q, from largest to smallest.

a) a, b, c b) b, c, a c) c, b, a d) All the m/q ratios are equal

21/08/2013

Quick Quiz



Now, rank the particles that arrive at a, b and c by their mass to charge ratio, m/q. c) c, b, a We can not determine individual masses or charges, but we can rank the particles in m/q. The particles that go through the largest circle must have the largest m/q ratio…

21/08/2013

Quick Answer



Most practical applications of electricity deal with electric currents – flows of charge from one region of space to another. Electric current is the rate of flow of charge through some region of space. The SI unit of current is the ampere (A). 1 A = 1 C / s The symbol for electric current is I. 21/08/2013

Electric Current



A force is exerted on a current-carrying wire placed in a magnetic field. The current is a collection of many charged particles in motion. The direction of the force is given by the right hand rule…

21/08/2013

Force on a Current Carrying Conductor



In the figure to the right, no current is flowing through the wire. Since no charge is moving within the wire, no force is exerted on it by the magnetic field. The wire therefore remains vertical.

21/08/2013

Force on a Wire



In the figure to the right, a current is flowing through the wire from the bottom to the top. By definition, current flows from positive to negative (even though normally it is carried by electrons, which flow from negative to positive!). As such, the direction current flows in is the direction that a positive charge would move. Therefore, the right hand rule shows us that the wire deflects to the left.

21/08/2013

Force on a Wire



In the figure to the right, current is now flowing through the wire from the top to the bottom. The force is therefore to the right (right hand rule, again). The wire therefore deflects to the right.

21/08/2013

Force on a Wire



Magnetic force is exerted on each moving charge inside the wire. That force is given by:

𝑭 = 𝑞𝒗𝒅 × 𝑩 The total force is the product of the force on one charge and the number of charges, i.e.:

𝑭 = 𝑞𝒗𝒅 × 𝑩 𝑛𝐴𝐿 where n is the number of charges per unit volume, A is the cross sectional area of the wire, and L the length of the segment. 21/08/2013

Force on a Wire: The Maths



𝑭 = 𝑞𝒗𝒅 × 𝑩 𝑛𝐴𝐿

Now the current, I, is given by the number of charges (of charge q) flowing through area A with velocity vD. We can therefore simplify the equation by substituting in that

𝐼 = 𝑛𝑞𝐴𝑣𝐷 To give us:

𝑭 = 𝐼𝑳 × 𝑩 where L is a vector that points in the direction of the current, and whose magnitude is the length, L of the segment.

21/08/2013

Force on a Wire: The Maths



Consider a small segment of the wire, ds . The force exerted on that segment is:

𝑑𝑭𝑩 = 𝐼 𝑑𝒔 × 𝑩 To work out the total force, we just integrate between the points of interest, i.e.:

𝑭𝑩 = 𝐼 𝑑𝒔 × 𝑩𝑏

𝑎

21/08/2013

Force on a Wire of Arbitrary Shape



Suppose that the field, B, is uniform.

Our integral 𝑭𝑩 = 𝐼 𝑑𝒔 × 𝑩𝑏

𝑎

Becomes 𝑭𝑩 = 𝐼𝑳

′ × 𝑩 (since the integral from a to b of ds is the vector sum of all length elements from a to b – i.e. the line between a and b). The magnetic force on a curved current carrying wire in a uniform field is equal to that on a straight wire connecting the end points and carrying the same current.

21/08/2013

Force on a Wire: Case 1



Again, consider a uniform field, B.

Our integral 𝑭𝑩 = 𝐼 𝑑𝒔 × 𝑩𝑏

𝑎

This becomes 𝑭𝑩 = 𝐼 𝑑𝒔 × 𝑩 = 0

The length elements here form a closed loop, so their vector sum is zero. Therefore, the net magnetic force acting on any closed current loop in a uniform magnetic field is zero.

21/08/2013

Force on a Wire: Case 2



The four wires shown to the right all carry the same current from point A to point B through the same magnetic field. Which of the following choices ranks them according to the magnitude of the magnetic force exerted on them, from the greatest force, to the least? a) a, b, c

b) a, c, b

c) a, c, d

d) a, d, c

e) b, c, d

f) c, b, d

21/08/2013

Quick Quiz



Which choices ranks them according to the magnitude of the magnetic force exerted on them, from the greatest force, to the least?

c) a, c, d

The order is (a), (b) = (c), (d).

The magnitude of the force depends on the value of sin θ. The maximum force occurs when the wire is perpendicular to the field – i.e. case (a). There is zero force in case (d), when the wire is parallel to the field.

Choices (b) and (c) represent the same force because a straight wire between A and B will have the same force on it as the curved wire, for a uniform field.

21/08/2013

Quick Answer



motion in a uniform magnetic field continued force on a...

Documents