| ⇦ | 
| ⇨ |
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 wavelength of kα, X-rays produced by an X-ray tube 0.76Å. The atomic number
- Dimensional formula of magnetic field in terms of mass M, length L
- A particle free to move along X-axis has potential energy given as U(X) =k(1-e⁻ˣ²)
- The identical cells connected in series are needed to heat a wire of length one meter
- What is the nature of Gaussian surface involved in Gauss’s law of electrostatics?
Topics: Magnetic Effects of Current and Magnetism
(167)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The wavelength of kα, X-rays produced by an X-ray tube 0.76Å. The atomic number
- Dimensional formula of magnetic field in terms of mass M, length L
- A particle free to move along X-axis has potential energy given as U(X) =k(1-e⁻ˣ²)
- The identical cells connected in series are needed to heat a wire of length one meter
- What is the nature of Gaussian surface involved in Gauss’s law of electrostatics?
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