| ⇦ |
| ⇨ |
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: - Two particles P and Q start from origin and execute simple harmonic motion along X-axis
- A particle of mass 100 g is thrown vertically upwards with a speed of 5 m/s. The work
- A gas is suddenly expanded such that its final volume becomes 3 times
- In a parallel plate capacitor of capacitance C, a metal sheet is inserted
- A thin prism of angle 15⁰ made of glass of refractive index µ₁ = 1.5 is combined
Topics: Magnetic Effects of Current and Magnetism
(167)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Two particles P and Q start from origin and execute simple harmonic motion along X-axis
- A particle of mass 100 g is thrown vertically upwards with a speed of 5 m/s. The work
- A gas is suddenly expanded such that its final volume becomes 3 times
- In a parallel plate capacitor of capacitance C, a metal sheet is inserted
- A thin prism of angle 15⁰ made of glass of refractive index µ₁ = 1.5 is combined
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