“There is only one corner of the universe that you can be certain of
improving, and that’s your own self.”
– Aldous Huxley
If a charged particle is projected at right angles to a magnetic
field, it follows a circular path of radius ‘r’ which is obtained by equating the magnetic force on it to the
centripetal force. Thus we have
qvB = mv2/r
[qvB is the magnitude of the magnetic
force on the particle of charge q moving
with speed v at right angles to the
magnetic field of flux density B and mv2/r is the centripetal force on the particle of mass m when it moves along a circle of radius
r].
From this we get r = mv/qB.
The time taken by
the particle to travel once round the circle is the period T of the circular motion.
We have T = 2πr/v
= 2πm/qB. The reciprocal of this (qB/2πm)
is called the cyclotron frequency.
If the
direction of projection of a charged particle is parallel or anti parallel to a
magnetic field, it continues to move along its straight line path since there is
no magnetic force on it. In general, if a charged particle is projected at an
angle θ with respect to a magnetic
field, it travels along a helical
path because the component (v sinθ) of its velocity perpendicular to the
field makes it move along a circle and the component (v cosθ) of velocity
parallel to the field makes it move along the field direction. The radius ‘r’ of the helix is obtained by equating
the magnetic force to the centripetal force.
Thus we
have
q(vsinθ)B
= m(vsinθ)2/r
Therefore,
r
= m(vsinθ)/qB
The
period of the circular component of motion while moving along the helix is the
same as that while moving along a pure circle.
Thus period = 2πr/vsinθ = 2πm/qB,
on substituting for ‘r’. Note that
the time taken by the charged particle to travel once round the circle (in the
case of circular path) is the same as the time of travel along one loop
of the helix (in the case of helical path) and this is independent of the
velocity of the particle.
Pitch of
the helical path = Parallel component of velocity× period= vcosθ (2πm/qB).
Now, consider the following
questions:
(1) A magnetic field of flux density B along the
Y-direction exists in a region of space contained between the planes x = a and
x = a + b. The minimum velocity with which a particle of mass ‘m’ and charge
‘q’ should be projected in the X-direction so that it can cross the magnetic
field is
(a) qB(a+b)/m
(b) qB/ma
(c) qB/mb
(d) qBa/m
(e) qBb/m
The
direction of projection of the particle is perpendicular to the magnetic field.
In the magnetic field, the particle moves along a circular path of radius r = mv/qB.
On increasing the velocity, the radius of the path increases and just before
crossing the field, the particle moves along a semi circle of radius ‘b’ within
the field. In this case we have, b = mv/qB so that v = qBb/m. When the velocity
exceeds this value, the particle crosses the magnetic field. The correct option
therefore is (e).
(2) A proton (mass 1.67×10–27 kg) is projected with a velocity of 4×106m/s
at an angle of 30º with respect to a uniform magnetic field of flux density 0.4
tesla. The path of the proton within the magnetic field is
(a) a circle of radius 2.5cm
(b) a spiral of radius 2.5cm
(c) a spiral of radius 5cm
(d) a spiral of radius 10cm
(e) a circle of radius 5cm.
The path
is a spiral (helix) since the direction of projection is not at right angles to
the magnetic field.
The
radius of the spiral is r = mvsinθ/qB = 1.67×10–27×4x106×(½) / 1.6×10–19×0.4
= 0.05 m = 5 cm. So, the correct option is (c).
You can find some useful questions
with solution here.
