Magnetic effect of current part-II : Magnetic effect of current part-II By R.Buvana
Lorentz’s Force : Lorentz’s Force Magnetic Lorentz Force acting on moving charge is 1.directly proportional to the magnitude of charge
2.directly proportional to the strength of magnetic field
3.directly proportional to component of velocity perpendicular to the magnetic field
F=qBv sin? and its direction may also be found by Fleming’s left hand rule.
Total Lorentz Force : Total Lorentz Force F=qE
F=qBv sin?
F(total)=qE+qBv sin?
Motion of a charged particle in E.F : Motion of a charged particle in E.F E=V/d, t=x/v, F=q E, a=q E/m
The distance y covered by the charged particle is
y= u t +1/2 at² =o x t +1/2 q E/m (x/v)²
y= q E/2mv² x².
This is the equation of a parabola and hence inside the electric field, a charged particle moves along the curved path ,which is parabolic in nature.
Motion of a charged particle in M.F : Motion of a charged particle in M.F When a charged particle having charge q moves inside a magnetic field B with velocity v it experiences a force given by
F=q v B sin?.Since v and B are at right angles to each other, Bqv=mv²/r.
r= m v/B q, T=2pm/B q, ?=B q/m.
From equations it follows that the period of the circular motion and the angular frequency of a charged particle moving in a uniform magnetic field neither depends upon the magnitude of its velocity nor upon the radius of the circular path. This fact is made use of to accelerate positive ions and such an accelerator is called cyclotron.
Cyclotron : Cyclotron The cyclotron devised by Lawrence and Livingston, is a device for accelerating ions to high speed by the repeated application of accelerating potentials.
Principle-It is based on the principle that a positive ion can acquire sufficiently large energy with a comparatively smaller alternating potential difference by making it to cross the same electric field time and again by making use of a strong magnetic field.
Conti… : Conti… Resonance condition- The condition of working of cyclotron is that the frequency of radio frequency alternating potential must be equal to the frequency of revolution of charged particle within the dees. This is called resonance condition.
Conti… : Conti… Suppose the positive ion with charge q moves in a dee with a velocity v,
qvB=mv²/r
r=mv/qB
t=p/?=pm/qB T/2=t= pm/qB
T=2pm/Bq
?=1/T=1/2pm/Bq v=qBR/m
K.E.=1/2 mv²=q²B²R²/2m
Limitations of the cyclotron : Limitations of the cyclotron When charged particle crosses gap of dees it gains K.E.=qV
Limitations of the cyclotron-
1.The energy of charged particles emerging from the cyclotron is limited due to the variation of mass with velocity, given by m=m0/(1-v²/c²)½ obviously the frequency of rotation of the charged particle decreases with increase of velocity .As a result it takes a longer time to complete semicircular path and continuously goes on lagging behind the alternating potential difference till a stage is reached when it can no longer be accelerated further.
Conti… : Conti… 2.Electrons are not accelerated by cyclotron-cyclotron is not used to accelerate electrons because electrons soon get so much velocity that their mass begins to increase with velocity and so their frequency becomes lower than frequency of radio frequency alternating voltage and the electrons revolve out resonance and hence their energy ceases to increase.
Force on a current carrying conductor : Force on a current carrying conductor F=BIl sin?
If the whole length of the conductor is perpendicular to the magnetic field ?=90° then sin90° =1, the force will be maximum, given by F=BIl
Force between two parallel current carrying conductors : Force between two parallel current carrying conductors F=µo/4 p 2I1 I2/r
According to Fleming’ left hand rule, the direction of magnetic force will be attractive. Other hand if the currents I1 and I2 in wires are in opposite directions, the force will be repulsive the magnitude of force in each case remains the same. If I1 =I2=1A,r=1m,then F=2x10¯7N / m. Thus one ampere is that current , which when flowing through each of the two parallel conductors of infinite length and placed in free space at a distance on one metre from each other, produces between them a force of 2 x 10¯7 newton per metre of their lengths.
Homework questions : Homework questions 1.A wire 12 cm long and carrying a current of 2A is placed perpendicular to a uniform magnetic field. If a force of 0.8N acts on it, calculate the value of the magnetic induction. (3.33T)
2.Why does a solenoid contract when current passes through it?
3.A current of 10A flows through each of two parallel long wires.The wires are 5 cm apart. Calculate the force acting per unit length of each wire. (4 x10?4 N/m)
Conti : Conti Answer 2.Current flowing in different turns, which are parallel to each other is in the same direction resulting into attraction between them and thus contraction of the solenoid.