Question

A particle with a charge of 5.0 µC and a mass of 5.0 × 10−12 kg is projected with a speed of 1.0 km s−1 in a magnetic field of magnitude 5.0 mT. The angle between the magnetic field and the velocity is sin−1 (0.90). Show that the path of the particle will be a helix. Find the diameter of the helix and its pitch.

Answer 1

Answer:

34. Magnetic Field

Answer :

Given-

Charge on the particle, q = 5 μC = 5 × 10−6 C

Magnetic field, B = 5 × 10−3 T

Mass of the particle, m = 5 × 10−12 kg

Velocity of projection, v = 1 Km/s = 103 m/s

Angle between the magnetic field and velocity,

θ= sin−1(0.9)

Since there are no forces in the horizontal direction ie there is no force in the direction of magnetic field, so, the particle moves with uniform velocity in horizontal direction.

Thus, it moves in a helical form.In helical motion we have two components of velocity.

Component of velocity which is perpendicular to the magnetic field is given by -

Similarly component of velocity in the direction of magnetic field will be the parallel component given by -

The velocity has a vertical component along which it accelerates with an acceleration a and moves in a circular cross-section.

We know motion in helical direction which is centripetal force and is given by –

Now, this force is balanced by Lorentz force acting due to presence of magnetic field ,

HOPE IT HELPED

Answer 2

Explanation:

Particle’s charge,$q=5 \times 10^{-6} \mathrm{C}$

Intensity of the magnetic field, $B=5 \times 10^{-3} \mathrm{T}$

Particle’s mass, $m=5 \times 10^{-12} \mathrm{kg}$

Projection velocity, $v=10^{3} \mathrm{m} / \mathrm{s}$

Angle between the velocity and the magnetic field, $\theta=\sin ^{-1}(0.9)$

Perpendicular to the magnetic field, the component of velocity is $v \perp=v \sin \theta$.

The magnetic field’s direction’s component of velocity, $v \|=v \cos \theta$.

In the horizontal direction, there are no forces and there will be the movement of the particle with uniform velocity.

A vertical component is seen in the velocity along with an acceleration a. It moves in a rounded cross-section. So, it travels in a helix.

$m v \perp 2 r=q$

$v \perp B$

$\Rightarrow r=0.18 m$

Hence, the helix diameter, $2 r=36 c m$

Pitch,

$P=2 \pi r v \sin \theta \times v \cos \theta=55 \mathrm{cm}$ .