Question

An alpha particle is revolving in the anticlockwise direction in a circular orbit of with a radius of R with a velocity or 'o' The orbit is such that the normal to the plane of the orbit makes an angle of 37° with the magnetic field B. The magnitude of torque experienced by the particle is:- ​

Answer 1

question

An alpha particle is revolving in the anticlockwise direction in a circular orbit of with a radius of R with a velocity or 'o' The orbit is such that the normal to the plane of the orbit makes an angle of 37° with the magnetic field B. The magnitude of torque experienced by the particle is:-

Explanation:

answer

in the above attachment

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Answer 2

The Torque Experienced By The Particle Is

$\frac{3ehB}{20\pi \: m} Nm$

GIVEN

The radius of the circular orbit = R

The velocity of the particle = o

Angel made by normal to plane with magnetic field = 37°

TO FIND

Torque is experienced by the particle.

SOLUTION

We can simply solve the above problem as follows-

We know that,

In the ground state,

Principle quantum number, n = 1

According to Bohr's theory, the angular momentum of the particle is given by,

$mvr = \frac{nh}{2\pi}$

where,

m = mass

v = velocity

r = radius

Putting the values in the above equation, we get,

$v = \frac{h}{2m\pi \: R}$

We know that,

time = distance/speed.

where,

Time = time taken

distance = circumference of the orbit

Putting the values in the above formula, we get

$T = \frac{2\pi \: R}{v \: }$

Putting the value of 'v';

$= \frac{2\pi \: R}{ \frac{h}{2\pi \: mR} }$

$= \frac{4 {\pi}^{2}m {R}^{2} }{h}$

We know that,

The magnetic moment of the particle is the product of the area and current flowing.

So,

M = in

we can also write it as ;

$M = \frac{charge}{time \: period} \times \pi {r}^{2}$

$M = \frac{e}{\frac{4 {\pi}^{2}m {R}^{2} }{h} } \times \pi {R}^{2}$

$m = \frac{eh}{4\pi \: m}$

The direction of the magnetic moment-

τ = M x B

|τ|

= MBsinθ

Putting the value of M in the above formula we get,

$τ = \frac{eh}{4\pi \: m \: } \times B \sin(37)$

$τ = \frac{3ehB}{20\pi \: m \: }$

Hence, the torque experienced by the particle is

$\frac{3ehB}{20\pi \: m} Nm$

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