“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 = mv*

^{2}/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*mv*^{2}*/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 =*2*πr/v =*2*πm/qB*. The reciprocal of this (*qB/*2*πm*) 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*sin*θ*) of its velocity perpendicular to the field makes it move along a circle and the component (*v*cos*θ*) 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*(

*v*sin

*θ*)

*B*=

*m*(

*v*sin

*θ*)

^{2}/

*r*

Therefore,

*r = m*(*v*sin*θ*)*/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 = 2

*πr/v*sin*θ*= 2*πm/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=

**.**

*v*cos*θ*(2*πm/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 4×10^{6}m/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}×4x10^{6}×(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.