Question

An alpha particle travelling at the velocity of 100 km/s flies at an angle 20° into a 1 meter wide area of homogeneous magnetic field with induction 5 mT as shown in the figure. The velocity direction of alpha particle is perpendicular to the field induction lines. In what time will the alpha particle leave the homogeneous magnetic field? Neglect the gravity action. Take the alpha particle charge-to mass ratio as 4.8 x (10)^7 C/kg. Give your answer in microseconds and round it to the whole.

Answer 1

Magnetic field is in Z direction. Angle rotated by velocity vector of alpha particle is $\theta$ and radius of circle made by particle in field is R. Width of region of magnetic field is 1meter.

$\vec B=B \hat{z}$ where B=5mT, Velocity of alpha particle is 100km/sec.

Magnetic force on charge particle:

$$F_m=q(\vec{V}\times\vec{B})\\ F_m=qVB......Eq[1]$$

Centripetal force on charge particle:

$$F_c=\cfrac{MV^2}{R}......Eq[2]$$

From Eq[1] & Eq[2],

Radius of circle of moving charge particle in a perpendicular magnetic field is:

$$R=\cfrac{MV}{qB}$$

Charge to mass ratio is given the question,

$$\cfrac{q}{M}=4.8\times{10^7}$$

Using all required values in expression of R, we get

$$R= \cfrac{5}{12}meter$$

since 2R< 1 meter (width of Magnetic field region), particle will return back to the first region.

From geometry we can calculate the angle $\theta=320\degree$, which in radian is $\cfrac{16\pi}{9}.$ Circular distance travelled by particle in filed is $S=2\pi R\times {\cfrac{\theta}{2\pi}}=\cfrac{16\pi R}{9}$ .

Time of travel inside the field is

$$T=\cfrac{S}{V}$$

putting all values of S and V we get

$$T=2.32\times 10^{-5} sec$$

Alpha particle will leave Magnetic field in time $23\mu.sec$ .