If a charged particle of 2 C moves with a velocity of 3 m/s perpendicular to a magnetic fieldof 1 T, then the magnetic force acting on the particle will be
Correct answer: D. 6 N.
- A. 3 N.
- B. 4 N.
- C. 5 N.
- D. 6 N.
Explanation
F = QvBsin(θ)Where: * F is the magnetic force * Q is the charge of the particle * v is the velocity of the particle * B is the magnetic field strength * θ (theta) is the angle between the velocity vector and the magnetic field vectorFrom the provided image, we have the following values: * Charge (Q) = 2 C * Velocity (v) = 3 m/s * Magnetic field (B) = 1 T * The particle moves perpendicular to the magnetic field, so the angle (θ) = 90°. * sin(90°) = 1Now, let's substitute these values into the formula:F = (2 C) * (3 m/s) * (1 T) * sin(90°)F = 2 * 3 * 1 * 1F = 6 N
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About Charged Particle in a Magnetic Field
A charged particle moving through a magnetic field experiences the Lorentz force, which is perpendicular to both its velocity and the field. The motion may be circular or helical, with questions involving radius, angular frequency, time period, and work done. A stationary charge feels no magnetic force.
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