<< Chapter < Page Chapter >> Page >

Problem : A particle carrying a charge 1μC is moving with velocity 3 i – 3 k in a uniform field -5 k . If units are SI units, then determine the angle between velocity and magnetic field vectors. Also determine the magnetic force.

Solution : The cosine of the enclosed angle is :

cos θ = v . B | v | | B | = 3 i 3 k . - 5 k | 3 i 3 k | | - 5 k | cos θ = 15 15 2 = 1 2

θ = 45 °

Magnetic force

Magnetic force is perpendicular to plane formed by velocity and magnetic field vectors.

The velocity and magnetic field vectors lie in x-z plane. The magnetic force is :

F M = q v X B = 1 X 10 - 6 [ 3 i 3 k X - 5 k ] F M = 1 X 10 - 6 X 15 j = 15 X 10 - 6 j

Magnetic force is along positive y – direction, which is perpendicular to the x-z plane of velocity and magnetic field vectors.

Context of electromagnetic interactions

In the discussion so far, we have assumed existence of electrical and magnetic fields. Here, we shall consider about the manner in which electrical and magnetic fields are set up by a source like charge or current and then investigate forces being experienced by the test charge. We shall consider three important cases in which (i) a stationary charge sets up an electrical field (ii) a moving charge sets up both electrical and magnetic fields and (iii) a current carrying wire sets up magnetic field. For each case, we shall discuss two states of test charge (i) it is stationary and (ii) it is moving. Also, note that we shall be deliberately concentrating on the forces experienced by the test charge. It is, however, implied that source charge or conductor carrying current also experiences the same amount of force in accordance with Newton’s third law of motion.

Force due to stationary charge

A stationary point source charge changes electrical properties of space around it. This property is quantified by the electrical field E at a particular point. If another point test charge is brought at that point, then it experiences electrical force, which is given by electrical part of the Lorentz force.

What happens when the test charge is moving also? It still experiences only the electrical force. No magnetic force is in play. See here that stationary source charge produces only electrical field around it. On the other hand, moving charge brought in its field sets up both electric and magnetic fields. The electric field is set up because moving test charge represents a net charge. But since it is also moving, magnetic field is set up by it in its surrounding in accordance with Biot-Savart Law.

We can easily see that two electrical fields (one due to stationary source charge and other due to moving test charge) interact to result in electrical force. However, there is only one magnetic field due to moving test charge without other magnetic field to interact with. As such, moving charge experiences only Lorentz electrical force in the presence of a stationary source charge.

Force due to moving charge

We now consider a moving charge, which acts as the source for setting up the fields. A moving charge produces both electrical and magnetic fields. If we bring another charge in its surrounding, then it experiences only electrical force. No magnetic force is in play. A stationary test charge only produces electrical field. There is no magnetic field to interact with the magnetic field produced by the moving source charge.

However, if we introduce moving test charge in the surrounding of source moving charge, then the moving test charge experiences both electrical and magnetic fields except for the situation when motion of the charge is neither parallel or anti-parallel to the magnetic field. However, if the motion of test charge is either parallel or anti-parallel to magnetic field produced by moving source charge, then the test charge only experiences electrical force.

Force due to current in wire

The current in wire sets up magnetic field in accordance with Biot-Savart law. Importantly, it does not set up electric field around it. Current through conductor is equivalent to passage of charge. Though, there is net transfer of electrons across a cross section of wire, but there is no accumulation of charge anywhere. As such, the wire carrying current is charge neutral even though there is flow of charge through it.

Now when a test charge is brought at a point in the surrounding of wire, the test charge does not experience any force. The wire sets up a magnetic field whereas charge sets up electrical field. These two different field types do not interact and there is no force on the test charge. On the other hand, if test charge is moving with certain velocity then it sets up electrical as well as magnetic fields. Two magnetic fields interact and as a result, the test charge experiences magnetic force except for the situation when motion of the charge is either parallel or anti-parallel to the magnetic field of the current in wire.

Magnetic field (b)

Strangely we have discussed and used the concept of magnetic field quite frequently, but without even defining it. There are certain difficulties involved here. There is no magnetic monopole like electrical monopole i.e. point charge. The smallest unit considered to be the source of magnetic field is a small current element. The Biot-Savart law gives relation for magnetic field due to a small current element. But, it is not quantifiable. How much is the “small” magnetic field or the “small” current length element?

As a matter of fact, the expression of Lorentz magnetic force provides us a measurable set up which can be used to define magnetic field. We have noted that magnitude of magnetic force is maximum when angle between velocity and magnetic field vectors is right angle.

F max = q v B B = F max q v

Thus we can define magnetic field (B) as a vector whose magnitude is equal to the maximum force experienced by a charge q divided the product “qv”. The direction of magnetic field is given by vector expression q( vXB ). The SI unit of magnetic field is Tesla, which is written in abbreviated form as T. One Tesla (T), therefore, is defined as the magnetic field under which 1 coulomb test charge moving in perpendicular direction to it at a velocity 1 m/s experiences a force of 1 Newton.

Exercise

A proton is projected in positive x-direction with a speed of 3 m/s in a magnetic field of (2 i +3 j ) X 10 - 6 T . Determine the force experienced by the particle.

Here,

v = 3 i m / s B = 2 i + 3 j X 10 - 6 T q = 1.6 X 10 - 19 C

The magnetic force is given by :

F M = q v X B F M = 1.6 X 10 - 19 [ 3 i X 2 i + 3 j 10 - 6 ] F M = 1.6 X 10 - 19 X 9 X 10 - 6 k F M = 1.44 X 10 - 24 k Newton

Questions & Answers

a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
y=10×
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
is it 3×y ?
Joan Reply
J, combine like terms 7x-4y
Bridget Reply
im not good at math so would this help me
Rachael Reply
how did I we'll learn this
Noor Reply
f(x)= 2|x+5| find f(-6)
Prince Reply
f(n)= 2n + 1
Samantha Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
Got questions? Join the online conversation and get instant answers!
QuizOver.com Reply

Get the best Algebra and trigonometry course in your pocket!





Source:  OpenStax, Electricity and magnetism. OpenStax CNX. Oct 20, 2009 Download for free at http://cnx.org/content/col10909/1.13
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Electricity and magnetism' conversation and receive update notifications?

Ask