A magnetic field has a magnitude of 1.20 10-3 T, and an electric field has a magnitude of 5.80 103 N/C. Both fields point in the same direction. A positive 1.8-µC charge moves at a speed of 2.90 106 m/s in a direction that is perpendicular to both fields. Determine the magnitude of the net force that acts on the charge.

Given Information:

Magnetic field = B = 1.20x10⁻³ T

Electric field = E = 5.80x10³ N/C

Magnitude of charge = q = 1.8x10⁻⁶ C

speed of charge = v = 2.90x10⁶m/s

Required Information:

Net force on the charge = Fnet = ?

Net force on the charge = 0.0121 N

Explanation:

As we know  force due to electric field is given by

Fe = qE

Fe = 1.8x10⁻⁶*5.80x10³

Fe = 0.01044 N

As we know  force due to magnetic field is given by

Fm = qvB

Fm = 1.8x10⁻⁶*2.90x10⁶*1.20x10⁻³

Fm = 0.006264 N

The net force is given by

Fnet =√(Fe² + Fm²)

Fnet =√((0.01044)² + (0.006264)²)

Fnet = 0.0121 N

Therefore, the net force acting on the charge is 0.0121 N

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