Question

A point charge of 6.8 C moves at 6.5 × 104 m/s at an angle of 15° to a magnetic field that has a field strength of 1.4 T.

What is the magnitude of the magnetic force acting on the charge?

A) 1.6 × 10–1 N
B) 6.0 × 10–1 N
C) 1.6 × 105 N
D) 6.0 × 105 N

Answer 1

The correct answer is A) 1.6 x 10-1 N

Answer 2

As per the question, the point charge is given as [q] = 6.8 C

The velocity of the charged particle [v] =
`6.5*10^(4)\ m/s`

The magnetic field [B] = 1.4 T

The angle made between magnetic field and velocity
`[\theta]` = 15 degree.

We are asked to calculate the magnetic force experienced by the charged particle.

The magnetic force experienced by the charged particle is calculated as -

Magnetic force
`\vec F=q(\ \vec V * \vec B)`

i.e F =
`=qVBsin\theta`

`=6.8*6.5*10^(4)*1.4*sin15`

`=61.88*10^(4)sin15`

`=61.88*10^4*0.2588\ N`

`=16.016*10^4\ N`

`=1.6*10^5\ N`

Hence, the force experienced by the charged particle is C i.e
`1.6*10^5 N`