# 22.4 Magnetic field strength: force on a moving charge in a magnetic  (Page 2/7)

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## Calculating magnetic force: earth’s magnetic field on a charged glass rod

With the exception of compasses, you seldom see or personally experience forces due to the Earth’s small magnetic field. To illustrate this, suppose that in a physics lab you rub a glass rod with silk, placing a 20-nC positive charge on it. Calculate the force on the rod due to the Earth’s magnetic field, if you throw it with a horizontal velocity of 10 m/s due west in a place where the Earth’s field is due north parallel to the ground. (The direction of the force is determined with right hand rule 1 as shown in [link] .)

Strategy

We are given the charge, its velocity, and the magnetic field strength and direction. We can thus use the equation $F=\text{qvB}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta$ to find the force.

Solution

The magnetic force is

$F=\text{qvb}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta .$

We see that $\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =1$ , since the angle between the velocity and the direction of the field is $\text{90º}$ . Entering the other given quantities yields

$\begin{array}{lll}F& =& \left(\text{20}×{\text{10}}^{–9}\phantom{\rule{0.25em}{0ex}}C\right)\left(\text{10 m/s}\right)\left(5×{\text{10}}^{–5}\phantom{\rule{0.25em}{0ex}}T\right)\\ & =& 1×{\text{10}}^{\text{–11}}\phantom{\rule{0.25em}{0ex}}\left(C\cdot \text{m/s}\right)\left(\frac{N}{C\cdot \text{m/s}}\right)=1×{\text{10}}^{\text{–11}}\phantom{\rule{0.25em}{0ex}}N.\end{array}$

Discussion

This force is completely negligible on any macroscopic object, consistent with experience. (It is calculated to only one digit, since the Earth’s field varies with location and is given to only one digit.) The Earth’s magnetic field, however, does produce very important effects, particularly on submicroscopic particles. Some of these are explored in Force on a Moving Charge in a Magnetic Field: Examples and Applications .

## Section summary

• Magnetic fields exert a force on a moving charge q , the magnitude of which is
$F=\text{qvB}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta ,$
where $\theta$ is the angle between the directions of $v$ and $B$ .
• The SI unit for magnetic field strength $B$ is the tesla (T), which is related to other units by
$1 T=\frac{\text{1 N}}{C\cdot \text{m/s}}=\frac{\text{1 N}}{A\cdot m}.$
• The direction of the force on a moving charge is given by right hand rule 1 (RHR-1): Point the thumb of the right hand in the direction of $v$ , the fingers in the direction of $B$ , and a perpendicular to the palm points in the direction of $F$ .
• The force is perpendicular to the plane formed by $\mathbf{\text{v}}$ and $\mathbf{\text{B}}$ . Since the force is zero if $\mathbf{\text{v}}$ is parallel to $\mathbf{\text{B}}$ , charged particles often follow magnetic field lines rather than cross them.

## Conceptual questions

If a charged particle moves in a straight line through some region of space, can you say that the magnetic field in that region is necessarily zero?

## Problems&Exercises

What is the direction of the magnetic force on a positive charge that moves as shown in each of the six cases shown in [link] ?

(a) Left (West)

(b) Into the page

(c) Up (North)

(d) No force

(e) Right (East)

(f) Down (South)

Repeat [link] for a negative charge.

What is the direction of the velocity of a negative charge that experiences the magnetic force shown in each of the three cases in [link] , assuming it moves perpendicular to $\mathbf{\text{B}}?$

(a) East (right)

(b) Into page

(c) South (down)

Repeat [link] for a positive charge.

What is the direction of the magnetic field that produces the magnetic force on a positive charge as shown in each of the three cases in the figure below, assuming $\mathbf{\text{B}}$ is perpendicular to $\mathbf{\text{v}}$ ?

(a) Into page

(b) West (left)

(c) Out of page

Repeat [link] for a negative charge.

What is the maximum force on an aluminum rod with a $0\text{.}\text{100}\text{-μC}$ charge that you pass between the poles of a 1.50-T permanent magnet at a speed of 5.00 m/s? In what direction is the force?

$7\text{.}\text{50}×{\text{10}}^{-7}\phantom{\rule{0.25em}{0ex}}\text{N}$ perpendicular to both the magnetic field lines and the velocity

(a) Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a $0\text{.}\text{500}\text{-μC}$ charge and flies due west at a speed of 660 m/s over the Earth’s south magnetic pole, where the $8\text{.}\text{00}×{\text{10}}^{-5}\text{-T}$ magnetic field points straight up. What are the direction and the magnitude of the magnetic force on the plane? (b) Discuss whether the value obtained in part (a) implies this is a significant or negligible effect.

(a) A cosmic ray proton moving toward the Earth at $\text{5.00}×{\text{10}}^{7}\phantom{\rule{0.25em}{0ex}}\text{m/s}$ experiences a magnetic force of $1\text{.}\text{70}×{\text{10}}^{-\text{16}}\phantom{\rule{0.25em}{0ex}}\text{N}$ . What is the strength of the magnetic field if there is a $\text{45º}$ angle between it and the proton’s velocity? (b) Is the value obtained in part (a) consistent with the known strength of the Earth’s magnetic field on its surface? Discuss.

(a) $3\text{.}\text{01}×{\text{10}}^{-5}\phantom{\rule{0.25em}{0ex}}\text{T}$

(b) This is slightly less then the magnetic field strength of $5×{\text{10}}^{-5}\phantom{\rule{0.25em}{0ex}}\text{T}$ at the surface of the Earth, so it is consistent.

An electron moving at $4\text{.}\text{00}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{m/s}$ in a 1.25-T magnetic field experiences a magnetic force of $1\text{.}\text{40}×{\text{10}}^{-\text{16}}\phantom{\rule{0.25em}{0ex}}\text{N}$ . What angle does the velocity of the electron make with the magnetic field? There are two answers.

(a) A physicist performing a sensitive measurement wants to limit the magnetic force on a moving charge in her equipment to less than $1\text{.}\text{00}×{\text{10}}^{-\text{12}}\phantom{\rule{0.25em}{0ex}}N$ . What is the greatest the charge can be if it moves at a maximum speed of 30.0 m/s in the Earth’s field? (b) Discuss whether it would be difficult to limit the charge to less than the value found in (a) by comparing it with typical static electricity and noting that static is often absent.

(a) $6\text{.}\text{67}×{\text{10}}^{-\text{10}}\phantom{\rule{0.25em}{0ex}}\text{C}$ (taking the Earth’s field to be $5\text{.}\text{00}×{\text{10}}^{-5}\phantom{\rule{0.25em}{0ex}}\text{T}$ )

(b) Less than typical static, therefore difficult

Why is there no 2nd harmonic in the classical electron orbit?
how to reform magnet after been demagneted
A petrol engine has a output of 20 kilowatts and uses 4.5 kg of fuel for each hour of running. The energy given out when 1 kg of petrol is burnt is 4.8 × 10 to the power of 7 Joules. a) What is the energy output of the engine every hour? b) What is the energy input of the engine every hour?
what is the error during taking work done of a body..
what kind of error do you think? and work is held by which force?
Daniela
I am now in this group
smart
theory,laws,principles and what-a-view are not defined. why? you
A simple pendulum is used in a physics laboratory experiment to obtain an experimental value for the gravitational acceleration, g . A student measures the length of the pendulum to be 0.510 meters, displaces it 10 o from the equilibrium position, and releases it. Using a s
so what question are you passing across... sir?
Olalekan
Two masses of 2 kg and 4 kg are held with a compressed spring between them. If the masses are released, the spring will push them away from each other. If the smaller mass moves off with a velocity of 6m/s, what is the stored energy in the spring when it is compressed?
54 joule
babar
how?
rakesh
Reduce that two body problem into one body problem. Apply potential and k. E formula to get total energy of the system
rakesh
i dont think dere is any potential energy... by d virtue of no height present
Olalekan
there is compressed energy,dats only potential energy na?
rakesh
yes.. but... how will u approach that question without The Height in the question?
Olalekan
Can you explain how you get 54J?
Emmanuel
Because mine is 36J
Emmanuel
got 36J too
Douglas
OK the answer is 54J Babar is correct
Emmanuel
Conservation of Momentum
Emmanuel
woow i see.. can you give the formula for this
joshua
Two masses of 2 kg and 4 kg are held with a compressed spring between them. If the masses are released, the spring will push them away from each other. If the smaller mass moves off with a velocity of 6m/s, what is the stored energy in the spring when it is compressed? Asume there is no external force.
Emmanuel
Inuwa
By using the Quotient Rule dy/dx = 3y/(x +y)²
Emmanuel
3y/(x+y)²
Emmanuel
may be by using MC^2=MC^2 and Total energy=kinetic energy +potential energy so 1st find kinetic energy and den find potential energy which is stored energy
rakesh
i think i m correct
rakesh
But how?
Emmanuel
3y/(x+y)²
Douglas
what's the big bang?
yes what is it?
LamaBbake
it is the explanation of how the universe began
Zainab
yes
Ana
explain
Chinagorom
in
Chinagorom
it is a theory on how the universe began. to understand more I would suggest researching the topic online.
david
thanks guys
kwame
if a force of 12N is applied to load of 200g what us the work done
We can seek accelation first
Nancy
we are given f=12 m=200g which is 0.2kg now from 2nd law of newton a= f/m=60m/s*2 work done=force applied x displacement cos (theta) w= 12x60 =720nm/s*2
Mudang
this very interesting question very complicated for me, í need urgent help. 1,two buses A and B travel along the same road in the same direction from Harper city (asume They both started from the same point) to Monrovia. if bus A maintains a Speedy of 60km/h and bus B a Speedy of 75km/h, how many
mohammed
hours Will it take bus B to overtake bus A assuming bus B starts One hour after bus A started. what is the distance travelled by the buses when They meet?.
mohammed
pls í need help
mohammed
4000 work is done
Ana
speed=distance /time distance=speed/time
Ana
now use this formula
Ana
Julius
great Mudang
Kossi
babar
hey mudang there is a product of force and acceleration not force and displacement
babar
@Mohammed answer is 0.8hours or 48mins
Douglas
nice
A.d
its not possible
Olalekan
í want the working procedure
mohammed
the answer is given but how Will One arrive at it. the answers are 4hours and 300m.
mohammed
physics is the science that studies the non living nature
ancient greek language physis = nature
isidor
what is phyacs
if i am going to start studying physics where should i start?
I think from kinematics
Nancy
You can find physics books at the library or online. That's how I started.
Chelsea
And yes, kinematics is usually where you can begin.
Chelsea
study basic algebra and calculus and can start from classical mechanics
Mudang
yes think so but dimension is the best starting point
Obed
3 formula's of equations of motion
vf=vi+at........1 s=vit+1/2(at)2 vf2=vi2+2as
Ana
benjamin
those are the three .. what you wanna solve ?
Nihrantz
For first equation simply integrate formula of acceleration in the limit v and u
Tripti
For second itegrate velocity formula by ising first equation
Tripti
similarly for 3 one integrate acceleration again by multiplying and dividing term ds
Tripti
any methods can take to solve this eqtions
a=vf-vi/t vf-vi=at vf=vi+at......1
Ana
suppose a body starts with an initial velocity vi and travels with uniform acceleration a for a period of time t.the distance covered by a body in this time is "s" and its final velocity becomes vf
Ana
what is the question dear
Zeeshan
average velocity=(vi+vf)/2 distance travelled=average velocity ×time therefore s=vi+vf/2×t from the first equation of motion ,we have vf =vi+at s=[vi+(vi+at)]/2×t s=(2vi+at)/2×t s=bit+1/2at2
Ana
find the distance
Ana
how
Zeeshan
Two speakers are arranged so that sound waves with the same frequency are produced and radiated through a room. An interference pattern is created. Calculate the distance between the two speakers?
How can we calculate without any information?
Amir
I think the formulae used for this question is lambda=(ax)/D
Amir