Question

A particle (q = 5.0 nC, m = 3.0 μg) moves in a region where the magnetic field has components Bx = 2.0 mT, By = 3.0 mT, and Bz = −4.0 mT. At an instant when the speed of the particle is 5.0 km/s and the direction of its velocity is 120° relative to the magnetic field, what is the magnitude of the acceleration of the particle in m/s2?

Answer 1

Explanation:

It is given that,

Charge on the particle, $q=5\ nC=5\times 10^{-9}\ C$

Mass of the particle, $m=3\ \mu g=3\times 10^{-6}\ g=3\times 10^{-9}\ kg$

Magnetic field component, $B_x=2\ mT,B_y=3\ mT,B_z=-4\ mT$

Net magnetic field, $B=\sqrt{2^2+3^2+4^2}=5.38\ mT=0.00538\ T$

Speed of the particle, v = 5 km/s = 5000 m/s

Angle between velocity and magnetic field, $\theta=120$

Magnetic force is given by :

$F=qvB\ sin\theta$

$F=5\times 10^{-9}\times 5000\ m/s\times 0.00538\times sin(120)$

$F=1.16\times 10^{-7}\ N$

Acceleration of the particle is given by, $a=\dfrac{F}{m}$

$a=\dfrac{1.16\times 10^{-7}\ N}{3\times 10^{-9}\ kg}$

$a=38.6\ m/s^2$

So, the acceleration of the particle is 38.6 m/s². Hence, this is the required solution.