*Notes to a video lecture on http://www.unizor.com*

__Magnetism - Lorentz Force - Problems 1__

*Problem 1a*

Consider the experiment pictured below.

A copper wire (yellow) of resistance

*is connected to a battery with voltage*

**R***and is swinging on two connecting wires (green) in a magnetic field of a permanent magnet.*

**U**All green connections are assumed light and their weight can be ignored.

Also ignored should be their electric resistance. Assume the uniformity

of the magnetic field of a magnet with magnetic field lines directed

vertically and perpendicularly to a copper wire.

The mass of a copper wire is

*and its length is*

**M***.*

**L**The experiment is conducted in the gravitational field with a free fall acceleration

*.*

**g**The magnetic field exerts the Lorentz force onto a wire pushing it

horizontally out from the field space, so green vertical connectors to a

copper wire make angle

*with vertical.*

**φ**What is the intensity of a magnetic field

*?*

**B***Solution*

**T·cos(φ) = M·g**

**T = M·g/cos(φ)**

**F = T·sin(φ) = M·g·tan(φ)**

**I = U/R**

**F = I·L·B = U·L·B/R**

**M·g·tan(φ) = U·L·B/R**

**B = M·R·g·tan(φ)/(U·L)***Problem 1b*

An electric point-charge

*travels with a speed*

**q***along a wire of length*

**v***.*

**L**What is the value of the equivalent direct electric current

*in the wire that moves the same amount of electricity per unit of time?*

**I**What is the Lorentz force exerted onto a charge

*, if it moves in a uniform magnetic field of intensity*

**q***perpendicularly to the field lines with a speed*

**B***.*

**v***Solution*

Let

*be the time of traveling from the beginning to the end of a wire.*

**T**

**T = L/v**

**I = q/T = q·v/L**

**F = I·L·B = q·v·B**Notice, the Lorentz force onto a wire in case of only a point-charge

running through it does not depend on the length of a wire, as it is

applied only locally to a point-charge, not an entire wire. Would be the

same if a particle travels in vacuum with a magnetic field present.

*Problem 1c*

An electric point-charge

*of mass*

**q***enters a uniform magnetic field of intensity*

**m***perpendicularly to the field lines with a speed*

**B***.*

**v**Suggest some reasoning (rigorous proof is difficult) that the trajectory

of this charge should be a circle and determine the radius of this

circle.

*Solution*

The Lorentz force exerted on a point-charge

*, moving with speed*

**q***perpendicularly to force lines of a permanent magnetic field of intensity*

**v***, is directed always perpendicularly to a trajectory of a charge and equals to*

**B***(see previous problem).*

**F=q·v·B**Since the Lorentz force is always perpendicular to trajectory, the linear speed

*of a point-charge remains constant, while its direction always curves toward the direction of the force. Constant linear speed*

**v**

**v**means that the magnitude of the Lorentz force is also constant and only

direction changes to be perpendicular to a trajectory of a charge.

According to the Newton's Second Law, this force causes acceleration

*,*

**a=F/m**which is a vector of constant magnitude, since the Lorentz force has

constant magnitude and always perpendicular to a trajectory, since the

force causing this acceleration is always perpendicular to a trajectory.

So, the charge moves along a trajectory with constant linear speed and

constant acceleration always directed perpendicularly to a trajectory.

Every smooth curve at any point on an infinitesimal segment around this

point can be approximated by a small circular arc of some radius (

*radius of curvature*) with a center at some point (

*center of curvature*). If a curve of a trajectory on an infinitesimal segment is approximated by a circle of some radius

*, the relationship between a radius, linear speed and acceleration towards a center of this circle (*

**R***centripetal acceleration*), according to kinematics of rotational motion, is

**a = v²/R**Therefore,

**R = v²/a**Since

*and*

**v***are constant, the radius of a curvature*

**a**

**R**is constant, which is a good reason towards locally circular character

of the motion of a charge. It remains to be proven that the center of

the locally circular motion does not change its location, but this is a

more difficult task, which we will omit.

Hence,

**R = v²/a = m·v²/F =**

= m·v²/q·v·B = m·v/q·B= m·v²/q·v·B = m·v/q·B

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