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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.
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Topics: Magnetic Effects of Current and Magnetism
(167)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A beam of light of λ=600 nm from a distant source falls on a single slit 1 mm wide
- If the velocity of an electron increases, then its de Broglie wavelength will
- The fluctuation in the input voltage of 200 V to a domestic circuit is +10V
- In insulators (CB is Conduction Band and VB is Valence Band)
- The mass of a ⁷₃ Li nucleus is 0.042 u less than the sum of the masses of all its nucleons
Topics: Magnetic Effects of Current and Magnetism (167)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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Using v=EB,
we have , E = v B
E =(2×103)1.5
E =3×103NC−1