As a charged particle 'q' moving with a velocity `vec(v)` enters a uniform magnetic field `vec(B)`, it experience a force `vec(F) = q(vec(v) xx vec(B)). For theta = 0^(@) or 180^(@), theta` being the angle between `vec(v) and vec(B)`, force experienced is zero and the particle passes undeflected. For `theta = 90^(@)`, the particle moves along a circular arc and the magnetic force (qvB) provides the necessary centripetal force `(mv^(2)//r)`. For other values of `theta (theta !=0^(@), 180^(@), 90^(@))`, the charged particle moves along a helical path which is the resultant motion of simultaneous circular and translational motions.
Suppose a particle that carries a charge of magnitude q and has a mass `4 xx 10^(-15)` kg is moving in a region containing a uniform magnetic field `vec(B) = -0.4 hat(k) T`. At some instant, velocity of the particle is `vec(v) = (8 hat(i) - 6 hat(j) 4 hat(k)) xx 10^(6) m s^(-1)` and force acting on it has a magnitude 1.6 N
Motion of charged particle will be along a helical path with

Options

A. A translational component along x-direction and a circular component in the y-z plane
B. A translational component along y-direction and a circular component in the x-z plane
C. A translational component along z-axis and a circular component in the x-y plane
D. Direction of translational component and plane of circular component are uncertain

Answer

C

Explanation

The component of velocity in the direction of magnetic field does not produce any maganetic force so it cannot produce circular motion, it can only produce translation motion.
`vec(v)=(8 hat(i)-6hat(j)+4hat(k)) xx 10^(-6) m//s` and `vec(B)=0.4 hat(k)T`
so, `v_(z)= 4 hat(k)` will produce translation motion and `v_(x)=8 hat(i)` and `v_(y)=-6 hat(j)` will produce circular motion as they are perpendicular to the megnetic field.