| ⇦ |
| ⇨ |
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: - A toroid having 200 turns carries a current of 1 A. The average radius of the toroid
- A stationary particle explodes into two particles of masses m₁ and m₂ which move
- The periodic waves of intensities I₁ and I₂ pass through a region at the same time
- The energy required to charge a parallel plate condenser of plate separation d
- A thin semicircular conducting ring (PQR) of radius r is falling with its plane
Topics: Magnetic Effects of Current and Magnetism
(167)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A toroid having 200 turns carries a current of 1 A. The average radius of the toroid
- A stationary particle explodes into two particles of masses m₁ and m₂ which move
- The periodic waves of intensities I₁ and I₂ pass through a region at the same time
- The energy required to charge a parallel plate condenser of plate separation d
- A thin semicircular conducting ring (PQR) of radius r is falling with its plane
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