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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.
Related Questions: - Which one of the following statements is not correct in case of a semiconductor?
- The area of cross-section of one limb of an U-tube is twice that of the other.
- In insulators (CB is Conduction Band and VB is Valence Band)
- If the focal length of objective lens is increased, then magnifying power of
- A plane wave of wavelength 6250Å is incident normally on a slit of width 2×10⁻² cm.
Topics: Magnetic Effects of Current and Magnetism
(167)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Which one of the following statements is not correct in case of a semiconductor?
- The area of cross-section of one limb of an U-tube is twice that of the other.
- In insulators (CB is Conduction Band and VB is Valence Band)
- If the focal length of objective lens is increased, then magnifying power of
- A plane wave of wavelength 6250Å is incident normally on a slit of width 2×10⁻² cm.
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