Mod 6 Magnfild

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THE MAGNETIC FORCE AND FIELD. The magnetic force When the charges are in motion, the interaction them in no longer purely electric. An additional force called magnetic force comes into effect. The magnetic force between two charges q1 and q2, moving with velocities v1 and v2, is equal to F mag =

µ o q1 q 2 4π r 2

(

v 1 × v 2 × rˆ

)

….

….

….

….

(1)

where µo is called the permeability constant which is equal to 4 π x 10-7 W/Am and r is the distance between the two charges (see Figure 1.1). The ratio R of the magnetic force and the electric force is equal to

….. …. (2) Substituting the numerical values of ε0 and µo into eq.(2), the ratio R can be rewritten as

where c is the velocity of light in vacuum (c = 3 x 108 m/s). Clearly, the magnetic force is small compared with the electric force unless the speed of the particles is high (a significant fraction of the velocity of light).

Figure 1.1. Relevant vectors for definition of magnetic force.

The magnetic field Magnetic field surrounds and is produced by moving charged particle. The magnetic field around and within a current carrying conductor is set up by the current flowing through the conductor, where as the field outside and inside a magnetized body originates from the intra-atomic and intermolecular motion of charged particles in the body. The basic magnetic field vector B called the magnetic induction is defined in terms of magnetic lines of induction. In the following way The tangent to the lines of B at any point gives the direction of B at that point. The number of induction lines per unit area is proportional to the magnitude of the magnetic vector B.

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The flux ΦB for a magnetic field is given by the surface integral ΦB = ∫ B•d A …. ….. (3) A magnetic field B can be associated with the magnetic force. The magnetic field at some point can be determined by placing a test charge at that point and moving it with some velocity v. The test charge will experience a magnetic force FB.The magnetic field B is related to the magnetic force FB via FB = qov xB ….

….

….

(4)

Z

The magnitude of the magnetic deflecting force FB is given by FB = qov B sinθ

...

.......

FB

.......(5)

B

qo X

v

Where θ is the angle between v and B. FB being at right angles to the plane formed by v and B will always be at right angles either to v or B. Magnetic force vanishes as v = 0. Again magnetic force vanishes ifv is either parallel of anti parallel toB. the deflecting force has a maximum value when v is at right angles to B i.e. F⊥ = qovB The SI unit of magnetic induction is tesla (T) which is in turn termed as Weber / meter2 . A charge of one coulomb moving with a velocity of one meter/ second perpendicular to a magnetic field of one tesla experiences a force of one Newton. 1T= 1weber /meter2 = 1N-sec/coul-m = 1N/amp-m Another unit of B is gauss (G), which is given by 1T= 1weber /meter2 = 104 gauss {the fact that the magnetic force is always at right angles to the direction of motion means that the work done by this force on the particle in zero} If a charged particle moves through a region in which both an electric and a magnetic field are present the resultant force is found to be F = qoE + qov xB

[ Lorentz force]

…

…

…. (6)

Hall Effect Edwin Hall studied the deflection of conduction electron in a copper wire in 1879. This phenomenon of deflection of charge carrier in conductors by magnetic field is called Hall effect and is used to determine the type of charge carrier and their density.

Y

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Figure below shows a copper strip of width d, carrying a current i from the top of the figure to the bottom. The charge carriers are electrons and they drift from bottom to top. At the instant an external magnetic field pointing into the plane of the figure is just turned on, a magnetic deflecting force FB acts on each drifting electron pushing is toward the right edge of the strip. As time goes on electrons moving to the right mostly piling up on the right of the strip leaves uncompensated positive charge on the left edge. This separation of positive and negative charge creates an electric field within the strip pointing from left to right. This electric field exerts a force FE on each electron tending to push it to the left. An equilibrium quickly develops in which the electric force FE on each electron just balances the magnetic force FB (figure b). Then the drifting electrons move along the strip toward the top without further deflection toward the right edge and thus no further increase of the electric field E occurs. i × × ×B × × × × × ×

× × × × ×vd × × × ×

i

FB

× × × × × × × × ×

× × × × × × × × ×

× × × × × B × × × × × × × × × × × × ×

i

× × + × + × + vd × + × FE × + × + × +

E

FB -

× × × × × × × × ×

× × × × × × × × ×

× × ×B × × × × × ×

× × E × × FE × × - vd × × ×

×+ × ×+ ×+ FB ×+ × + × ×+ ×+

× × × × × × × × ×

(a) Magnetic force FB on electron (b) Electric force FE balances the FB (c) When free charge carrier is positive The hall potential difference associated with the strip across the strip of width d is VH = Ed …… ….. ……..……..(7) A voltmeter can measure this potential difference. Moreover a voltmeter can tell which edge is a higher potential. For electron the left edge is a higher potential. When the electric force and magnetic forces are in balance we can write eE = e vd B ……. ……. ……. ……. (8) J

i

The drift speed vd is related to current as vd = ne = n e A Where A is the area of cross section of the copper strip.

……. ……. (9)

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Combining equation (1) , (2) and (3) we get VH = vd Bd =

iBd …… ……..……… neA

……. (10)

iBd

or ,

n = V eA H

or ,

n=

iB VH t e

…..

……..……..……….

…… (11)

In which t (=A/d) is the thickness of the strip. Practical applications of Hall effect 1. Determination of the number density of charge carrier in conductors. 2. Determination of type of charge carriers from polarity of Hall potential drop 3. Determination of drift speed of charge carriers Circular motion of charged particle in uniform magnetic field Charged particle moving in a magnetic field feels a force perpendicular to their velocity. Since the force and hence acceleration of the charged particle is always perpendicular to their direction of motion the work done by magnetic force on it is zero. Hence its speed and kinetic energy remains unchanged. However its direction changes continuously resulting in a circular path. The radius of such a charged particle moving perpendicularly in a uniform magnetic field B is characterized by its mass m, speed v and charge q. Such a charged particle moving perpendicularly in uniform magnetic field experiences a magnetic deflecting force FB = q v B. From Newton’s second law applied to uniform circular motion we get mv2 Fc = m a = r

or ,

x

x

x

x

x

x

FB

m v2 r

mv r = ........ qB

x Fc

q

then we have q vB =

x

.........

(12)

x

x

x

x

x

x

x

x

x

x

x

x

The time period of revolution is equal to the circumference divided the speed T =

2π r 2π m v 2 πm = = v v qB qB

…… …. …….(13)

Hence the frequency qB 1 = . T 2 πm And the angular frequency qB ω = 2πf = m f =

.......

.....

(14)

........ .....

(15)

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Magnetic Force on a current As a magnetic field exerts a sideways force on a moving charge, it will exert a similar force on a wire carrying current. Let a wire of length l carrying current i be placed in a magnetic field of induction B as shown in the figure (2). For simplicity the current density j has been taken at right angles toB. The current is carried by free electrons in a conductor, n being the number of such free electron per unit volume of the wire. The magnitude of the average force on one such electron is FB = qov B sin θ = evd B …. J

…

…

i

Now vd = ne = n e A is the drift speed. Using the above equation  i   B FB = e n e A  

.

….

……

…

(16) x x

× ×

× F` × × ×

× x

x x

× ×

x x

Vd

× x

× ×

× ×

× ×

… (17) The length of the wire contains nAl free electrons as Al is the volume of the section of cross-section A. So the total force on the free electrons and thus on the wire itself is 

i   B = i l B ….  ne A

F = ( nAl ) e

……

…(18)

The above equation holds only if the wire is at right angles to B. The more general situation can be expressed in the vector form as F = il xB …. …… …… (19) The magnetic force on a differential current length element i dl is dF = i dl xB …. …… …… (20) Torque on a current loop Figure (3) shows a rectangular loop of wire of length a and width b placed in a uniform field of induction B, with sides cd and ef always normal to the field direction. The normal to the plane of the loop makes an angle a with the direction ofB. The loop is capable of rotating around an axis OO’. d

O’

F1

F3 i

α f c b sin α F4

F2 e

O

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Figure (a)

Figure (b)

The force on side cd is F1 = iaB …. …… …… (22) This force acts vertically upward normally to the directions of i and B. again the same amount of force F2 acts on side ef vertically downward. Again each of sides ce and df are inclined at α with the direction of B and are acted upon by forces of magnitude F’ = ibBsin α these tow forces have same line of action but are oppositely directed. So they do not have any net effect on the motion of the loop. On the other hand forces of equal magnitude and opposite direction acts on side cd and ef. The effect of these two force is to cause a rotation about OO’. The torque rotating the coil is τ′ = ia B bsin α …. …… …… (23) If there are N turns, then the torque on the entire coil is τ = Nτ’ = Nia Bb sin α Niab B sin α = Ni AB sin α …. …… …… (24) where A=ab is the area of the coil. The above relation can be written in the following form τ = µB sin α where m is the magnetic dipole moment of the loop. In the vector form τ =  µ X B …. …… …… (25) the magnetic potential energy isof the loop at any position q is defined as the work that an external agent must do to turn the dipole from its zero energy potion (q=900) to the given position q. Thus U=∫ τ dθ = NiAB sinθ dθ =µB sin θ dθ = -µB cosθ In vector symbol the above relation can be expressed in the following form U =- µ.B …. …… …… (26)

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Problem sheet : Magnetic field Magnetic force on charge 1. An alpha particle travels at a velocity of 550 m/s through a uniform magnetic field B of 0.045 T. Angle between v and B is 450 . (a) What is the magnitude of the force acting on the particle due to the field? (b) what is the acceleration of the particle due to this force? (c) Does he speed of the particle increase or decrease? [Charge of alpha particle q = 3.2 x 1019 C and mass mα = 6.6 x 10-27kg] 2. An electron has an initial velocity of (12.0 j + 15.0 k) km/sand a constant acceleration of (2.0 x 1012 m/s2)i in a region in which uniform electric and magnetic fields are present. If (B = 400 µT)i , find the electric field. x × × × × Hall effect x × × × x v 3 A metal strip 6.5 cm long, 0.85 cm wide × × × × × and 0.76 mm thick moves with constant x × × × × velocity v through a magnetic field B = 1.2 mT pointing perpendicular to the strip, as shown in the figure below. A .potential difference of 3.9 µV is measured between points x and y across the strip calculate the speed v. 4. A strip of copper 150 µm wide is placed in a uniform magnetic field B of magnitude 0.65 T , with B perpendicular to the the strip. A current I = 23 A is then sent through the strip such that a Hall potential difference V appears across the width. Calculate V. [Number of charge carriers per unit volume of copper is 8.47 x 1028 electrons / m3.] 5. In a Hall effect experiment, a current of 3.0 a sent length wise through a conductor 1.0 cm wide, 4.0 cm long and 10 µm thick produces a transverse (across the width) hall voltage of 10 µV when a magnetic field of 1.5 T is passed through the thickness of the conductor. From this data find (a) the drift velocity of charge carriers, (b) number density of charge carriers and(c) Show on a diagram the polarity of the Hall voltage with assumed current and magnetic field direction, assuming also that the charge carriers are electrons. 6. A metal strip 6.5 cm long, 0.85 cm wide and 0.76 mm thick moves with constant velocity v through a magnetic field B = 1.2 mT pointing perpendicular to the strip, as shown in the figure below. A potential difference of 3.9 µV is measured between points x and y across the strip calculate the speed v. Circulating Charge 7. An electron is accelerated from rest by a potential difference of 350 V. it then enters a magnetic field of magnitude 200 mT with its velocity

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perpendicular to the field. (i) Calculate the speed of the electron and (ii) the radius of the path in the magnetic field. 8. What uniform magnetic field, applied perpendicular to a beam of electrons moving at a speed of 1.3 x 106 m/s, is required to make the electrons travel in a circular arc of radius 0.35m? 9. An electron with kinetic energy 1.2 keV circles in a plane perpendicular the a uniform magnetic field. The orbit radius is 25.0 cm. Find (a) the speed of the electron, (b) the magnetic field , (c) the frequency of circling, and (d) the period of rotation. Magnetic force on a current carrying conductor 10 A horizontal conductor in a power line carries a current of 5000 A from south to north. Earths magnetic field (60 µT) is directed toward the north and is inclined downward at 700 to the horizontal. Find the magnitude and direction of force on 100 m of the conductor due to Earth’s field. 11 A wire of 62.0 cm length and 13.0 g × × × × x mass is suspended by a pair of flexible x leads in a magnetic field of 0.44 T as x × × × x x shown in figure below. What are the × × × × × x magnitude and direction of the current x × × × × x required to remove the tension in the supporting leads? 12 A wire 1.8 cm long carries a current of 13 A and makes an angle of 35 0 with a uniform magnetic field B = 1.5 T. calculate the force on the wire. 13. a metal wire of mass m slides m without friction on two horizontal rails G i spaced a distance d apart, as shown in figure below. The track lies is a vertical uniform magnetic field B. there is a constant current I through generator G, along one rail, across the wire, and back down the other rail. Find the speed and direction of the wire’s motion as a function of the time, assuming it to be stationary at t = 0