Physics, asked by indirasebastian1474, 6 months ago

A particle of charge -0.04C is projected with speed 2x10^4 m/s into a uniform magnetic field ‘B’ of strength 0.5T If the particle’s velocity as it enters field is perpendicular to ‘B’, what is the magnitude of the magnetic force on this particle?

Answers

Answered by pankajsbicif
1

Answer:

Magnetic force = 4×10^-6

Attachments:
Answered by Sahil3459
1

Answer:

The magnitude of the magnetic force on this particle will be 400 N.

Explanation:

Because of their velocity, electrically charged particles experience a magnetic force, attraction, or repulsion. It is the fundamental force that causes effects like the action of electric motors and the attraction of magnets to iron.

For above given question:

Magnetic force F will be = q(v × B)

As v⊥B ⟹∣F∣ = ∣q∣vB

Therefore, ∣F∣ = 0.04 × (2 × 10⁴) × 0.5 = 400 N

Thus, this also means that a charge that is stationary or traveling parallel to the magnetic field has no magnetic force.

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