Question

An electron with a velocity of 10^7 m/s enter into a region of magnetic flux density of 0.10T,the angle between the direction of the field and the initial path of electron being 25°.Find the axial distance between two turn of the helical path

Answer 1

Answer:

3.2 mm

Explanation:

Answer 2

As per the data given in above question

we have to find the axial distance or r

For any angle other than 0° ,90° and 180° the path is helical.

Now ,

1.The force for electron,F = q×v×B× sin (x ).....(i)

where x is angle

and

2.The force required to keep electron in circular path

F= mv²/r .........(ii)

from (i) and (ii) ,we get

$q \times v \times b \times sin(x) = \frac{m {v}^{2} }{r} ...(iii)$

solving equation (iii) ,we get

$r = \: \frac{m \times {v}^{2} }{q \times v \times b \times sin(x)}$

The terms are

$velocity\:or\: v = {10}^{7} \frac{m}{s}</p><p></p><p>magnetic\:flux \: or \: \: b = 0.10 \: tesla$

angle,x = 25º

sin(x)= sin(25°)= 0.4226

mass of electron ,m= 1.6×10-²⁷ kg

charge of electron,q= 1.6× 10-¹⁹ coulomb .

$r = \frac{1.6 \times {10}^{ - 27} \times {10}^{7} \times 0.4226}{1.6 \times {10}^{ - 19} \times 0.10 }$

$r = \frac{0.676 \times {10}^{ - 20} }{0.16 \times {10}^{ - 19} }$

$r = 0.42 \: metre$

Therefore,

The axial distance between two turn of the helical path is 0.42 metre