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A charged particle with velocity 2 x 103 m/s passes undeflected through electric and magnetic field. Magnetic field is 1.5 telsa. The electric field intensity would be
  • a)
    2x 103 N/C
  • b)
    1.5 x103 N/C
  • c)
    3 x 103 N/C
  • d)
    4/3 x 10⁻3N/C
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
A charged particle with velocity 2 x 103 m/s passes undeflected throug...
Given:
Velocity of the charged particle = 2 x 10^3 m/s
Magnetic field = 1.5 Tesla

To find:
Electric field intensity

Explanation:
When a charged particle moves through a magnetic field, it experiences a force called the magnetic force given by the equation:

F = qvBsinθ

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, and
θ is the angle between the velocity vector and the magnetic field vector.

In this case, the charged particle passes undeflected through the magnetic field, which means the magnetic force acting on it is zero. Therefore, the angle between the velocity vector and the magnetic field vector is 0 degrees.

sinθ = 0
θ = 0 degrees

Substituting the values in the equation for the magnetic force:

F = qvBsinθ
F = qvB(0)
F = 0

Since the magnetic force is zero, the electric field must balance out this force and keep the particle undeflected. The electric force is given by the equation:

F = qE

Where,
F is the electric force,
q is the charge of the particle, and
E is the electric field intensity.

Since the electric force is equal to zero, we can say:

F = qE
0 = qE

Therefore, the electric field intensity, E, must also be zero.

However, none of the given options match the correct answer. Therefore, there may be an error in the question or possible answers provided.
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Community Answer
A charged particle with velocity 2 x 103 m/s passes undeflected throug...
E/B=v
E=B×V
E=3×10^3
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A charged particle with velocity 2 x 103 m/s passes undeflected through electric and magnetic field. Magnetic field is 1.5 telsa. The electric field intensity would bea)2x 103 N/Cb)1.5 x103 N/Cc)3 x 103 N/Cd)4/3 x 103N/CCorrect answer is option 'C'. Can you explain this answer?
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