| ⇦ |
| ⇨ |
A charged particle with a velocity 2×10³ ms⁻¹ passes undeflected through electric field and magnetic fields in mutually perpendicular directions. The magnetic field is 1.5T. The magnitude of electric field will be
Options
(a) 1.5×10³ NC⁻¹
(b) 2×10³ NC⁻¹
(c) 3×10³ NC⁻¹
(d) 1.33×10³ NC⁻¹
Correct Answer:
3×10³ NC⁻¹
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - The rms current in an AC circuit is 2 A. If the wattless current be √3 A,
- Given the value of Rydberg constant is 10⁷ m⁻¹, the wave number of the last line
- In a uniform circular motion, work done in one complete rotation is
- If a magnetic substance is kept in a magnetic field, then which of the following
- If a small amount of antimony is added to germanium crystal
Topics: Magnetic Effects of Current and Magnetism
(167)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The rms current in an AC circuit is 2 A. If the wattless current be √3 A,
- Given the value of Rydberg constant is 10⁷ m⁻¹, the wave number of the last line
- In a uniform circular motion, work done in one complete rotation is
- If a magnetic substance is kept in a magnetic field, then which of the following
- If a small amount of antimony is added to germanium crystal
Topics: Magnetic Effects of Current and Magnetism (167)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends

Using v=EB,
we have , E = v B
E =(2×103)1.5
E =3×103NC−1