Question

Two charged particles are moving with equal velocities of 4.00 m/s in the +x-direction. At one instant of time the first particle with a charge of 7.40 μC is located at x = 0 and y = +1.60 cm, and the second particle with a charge of 3.20 μC is located at x = 0 and y = -1.60 cm.
1. What is the y-component of the magnetic force on the first particle due to the second?
2. How fast would the charges have to be moving for the magentic force to be equal in magnitude to the electric force?

Answer 1

Answer:

Explanation:

Electric  force between two particles

= ( 9x10⁹x7.4 x 10⁻⁶ x 3.2 x 10⁻⁶ )/  (3.2 x 10⁻²)² ( Distance between particles is 1.6 +1.6 = 3.2 cm . )

= 20.81 x 10 = 208.1 N

1 ) Magnetic field due to movement of second charge

= $\frac{\mu_0}{4\pi} \frac{qv}{r^2 }$

= 10⁻⁷ x 3.2 x 10⁻⁶ x 4 / (3.2 x 10⁻²)²

B = 1.25 x 10⁻⁹ T.

Due to this magnetic field , there is a force called magnetic force on first particle which will be expressed as follows

Force = Bqv

= 1.25 x 10⁻⁹ x 7.4 x 10⁻⁶ x 4

37 x 10⁻¹⁵ N

2 ) For magnetic force to be equal to electric force let velocity o particles be V

Then magnetic field due to second charge

= 10⁻⁷ x 3.2 x 10⁻⁶ x v / (3.2 x 10⁻²)²

= .3125 v x  10⁻⁹

Magnetic force on first charge

Bqv

=  .3125 v x  10⁻⁹  x 7.4 x 10⁻⁶ x v

2.3125 x 10⁻¹⁵ v²

magnetic force = electric force

2.3125 x 10⁻¹⁵ v² = 208.1

v² = 90 x 10¹⁵

v² = 900 x 10¹⁴

v = 30 x 10⁷ m /s