Question

Find the magnetic force experienced by an electronic of charge 1.6 x 10^-19 prejected into a magnetice field dentist iot , with a velocity of 3 x 10⁹ in a direction

Parallel time the field

At right angle to the field

At angle 30⁰ to the field

Answer 1

Given data

Charge is $q= 1.6\times10^{-19} C$

Velocity is $v=3\times 10^9 m/s$

Magnetic field is B (Magnetic field value not given in the question)

Case (I),

When the magnetic field is parallel to velocity, then angle between them is0^o . So the magnetic force is

$F=qvBsin0=0$

Case (II),

When the magnetic field is perpendicular to velocity, then angle between them is 90^o. So the magnetic force is

$F=qvBsin90^o=(1.60\times 10^{-19})(3\times 10^9)\times B$

Substitute value of B in the above equation, we can get the answer.

Case (III),

Angle between the magnetic field and velocity is 30^o . So the magnetic force is

$F=qvBsin30^o= \frac{1}{2}qvB$

Substitute values of q , v and B , we the magnetic force on the charged particle in magnetic field.

Note:As per question, the velocity of charged particle given as 3*10⁹ m/s. According to Einstein's relativity, No particle can move more than speed of light.