A charged particle with a velocity 2×103ms−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.
1.5×103NC−1
2×103NC−1
3×103NC−1
1.33×103NC−1
A
1.33×103NC−1
B
1.5×103NC−1
C
2×103NC−1
D
3×103NC−1
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Solution
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Given : v=2×103 m/s B=1.5T
As the particle passes undeflected, thus electric force is balanced by the magnetic force acting on the particle.
∴qE=q(v×B)
⟹E=Bv=1.5×2×103=3×103NC−1
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