Question

An alpha particle travels at a velocity of magnitude 520 m/s through a uniform magnetic field of magnitude 0.034 T. (An alpha particle has a charge of charge of +3.2 × 10-19 C and a mass 6.6 × 10-27 kg) The angle between the particle's direction of motion and the magnetic field is 43°. What is the magnitude of (a) the force acting on the particle due to the field, and (b) the acceleration of the particle due to this force? (c) Does the speed of the particle increase, decrease, or remain the same?

Answer 1

Answer:

(a) Magnetic force $F=38.584\times 10^{-19}N$

(b) Acceleration $a=5.846\times 10^8m/sec^2$

(C) Speed will remain same

Explanation:

We have given velocity of alpha particle v = 520 m/sec

Magnetic field B = 0.034 T

Charge on alpha particle $q=3.2\times 10^{-19}C$

Mass of alpha particle $m=6.6\times 10^{-27}kg$

Angle between velocity and magnetic field 43°

(a) Force acting on the particle is equal to

$F=q(v\times B)=qvBsin\Theta$

$F=3.2\times 10^{-19}\times 320\times 0.034\times sin43^{\circ}$

$F=38.584\times 10^{-19}N$

(B) According to newton's law

F = ma. here m is mass and a is acceleration.

So acceleration

$a=\frac{F}{m}=\frac{38.584\times 10^{-19}}{6.6\times 10^{-27}}=5.846\times 10^8m/sec^2$

(c) As the magnetic force is always perpendicular to velocity so speed will remain same neither decreases nor increases.