KPK Board 12th class Physics Ch 13 Electromagnetism short questions answers
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A current carrying conductor placed in a magnetic field experiences a magnetic force due to the interaction of the two magnetic fields. One is the field in which the conductor is placed and other is that magnetic field which produces around the conductor due to flow of current through it. The magnitude of this magnetic force is given by:
F = I(LxB)
F = ILBSinθn
F= ILBSinθ
Thus the magnetic force will be maximum when θ=90 i.e.
Conductor is placed perpendicular to the magnetic field. The magnetic force will be zero when θ=θ when conductor is parallel to the magnetic field.
Direction:
The direction of magnetic force is perpendicular to the both A and B and is determined by Fleming, s left hand rule.
The nature of the force between the two wires carrying current in the same direction is attractive. Because, each wires produce magnetic field around it due to flow of current. In this way each wire is placed in the magnetic field of the other as shown. The magnetic field B1 due to I1 according Ampere, s Law is
B1=µ0I1/2πr
The magnetic force on the write carrying current I2 is
F2 = I2LB = I2L[µ0I1/2πr]
F2 = µ0I1I2L/2πr
The direction of the magnetic field can be determined by Fleming’s left hand rule.
When a charged particle moves in a magnetic field of strength B, at an angle θ its own magnetic field interacts with the field B. As a result the magnetic force F experienced by the charged particle q is given by.
F = q[VxB]
F = qVBsinθn
F = qVBSinθ
This equation gives magnitude of the magnetic force which depends upon the angle θ in addition to the values of q, V and B.
- The magnetic force will be maximum at 90 degree.
- The magnetic field will be zero at zero degree.
When a charged particle moves with a velocity v enters a magnetic field B, it experience magnetic force. Therefore when a beam of electrons enter with a velocity v in a uniform magnetic field of strength B at angle θ with the field then each electron of charge e will experience magnetic force which is given by.
F = e[VxB]
F = eVBSinθn
F = eVBSinθ
- The magnetic force will be maximum at 90 degree.
- The magnetic field will be zero at zero degree.
When a charged particle q entering a uniform magnetic field of strength B with a velocity V at an angle θ to the magnetic field, it will move on a helical path as shown the velocity V of a particle is resolved into rectangular components.
Vx = Vcosθ
Vy = Vsinθ
- The component V Cos θ theta is parallel to the magnetic fields which cause the particle to move direction of B.
- V sin θ theta is perpendicular to B and cause the charged particle to move in a circle.
- Thus as a result the charged particle will move on a spiral path.
- The time period for 1 revolution is
T = 2πr/V
But qVB = mv2/r And V = qBr/m
So T = 2πm/qB
No, we cannot say with certainty that the magnetic field in a certain region of space is zero where a charged particle moves in a straight line, because a beam of electrons can travel straight through a certain region in two cases.
- When there is a electric and magnetic field in the region.
- When both electric and magnetic fields are there and acting perpendicular to each other. When their strength is so adjust that electric and magnetic force is balance each other as shown.
- Fm = F e
- evB = eE
V=B/E………..?
Thus when both B and C are kept constant then the beam of electron will move with constant velocity without any deflection.
An electron beam passing through a region of crossed electric field and magnetic field B will remain un-deflected, if the electron moves with constant speed. When the two fields are so adjust that electric and magnetic forces on the beam balance each other and net force on it is zero therefore,
Fm-Fe= 0
Fm = Fe
qVB = qE
eVB = eE
V = B/E
Thus when electric field and magnetic field B are both constant then the velocity of the beam will be constant and will not be deflected.
When an electron is projected in the direction of electric and magnetic field, its kinetic energy will be used to do work against the electric force.
Because the magnetic force on the electron is zero.
Fm = eVBsinθ
Fm = eVsinθ
Fm = 0
Thus the magnetic force does not affect the motion of electron.
Fe = -eE
Mα = – eE
Αlpha = -eE/m
As the electron is projected in the direction of electric field, so the electric force experienced by electron will be opposite to the electric field.
When Alpha charged particle Q moves in a magnetic field B, the magnetic force experienced by it provides the necessary centripetal force so that to move the charged particle in a circle. Therefore,
Fm = Fe
qVB = mv2/r
V = qBr/m
Where V is the speed of M is the mass of charged particle. Since we know that
V = rw
Rw = qBr/m
W = qB/m
W = 2f
2f = qb/m
F = qB/2m
Thus equation gives the cyclotron frequency of charged particle Q in magnetic field b.
No, neutron cannot be accelerated in a cyclotron, because they are neutral particles, they have no charge. A cyclotron is a charged particle acceleration machine. in a cyclotron the massive charged particle such as a proton and deuterons or Alpha particles are first accelerated through high potential difference such as 1000 volts, to get high energy then the magnetic field is applied to move the charged particle along a circular path. Neutrons are uncharged particle, which are not affected by electric field and magnetic field B. so they cannot be accelerated by cyclotron machine.
When a loop of a L sectional area, carrying current I is placed in a uniform magnetic field B, torque is experienced by the coil. If the planes of loop make an angle Alpha with the magnetic field B. then the torque is given by,
Torque = BIACosα
now if the loop is free to rotate, then it will rotate under the action of the above talk and Orient itself in the such a way that its plane becomes perpendicular to the magnetic field, the torque will be become zero and the law will not attend to rotate further,
Torque = BIACos90 = 0
Ans: A current carrying coil behave like a bar magnet. When a current carrying coil is suspended freely, the component of the Earth’s magnetic field being perpendicular to the plane of the coil will influence. The two magnetic fields on due to current and the other the Earth’s magnetic field will interact and exert magnetic force on the coil. As a result of this magnetic force, torque will exert on the coil which will rotate, so that the plane of the coil become parallel to the Earth’s magnetic field.
Since the direction of Earth’s magnetic field is along the north-south, therefore the plane of the current carrying coil will settle down along north south direction. Hence current carrying coil behaves like a bar magnet.
