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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
- The electric potential at a point (x,y,z) is given by V = – x²y – xz³ + 4.
- If the binding energy of the nucleon in ₃⁷Li and ₂⁴He nuclei are 5.60 MeV and 7.06 MeV
- In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series
- As per Bohr model, the minimum energy (in eV) required to remove an electron
- If the electric field lines is flowing along axis of a cylinder, then the flux
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