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A charged particle is moving on circular path with velocity v in a uniform magnetic field B, if the velocity of the charged particle is doubled and strength of Magnetic field is halved, then radius becomes
As
According to the question, v‘ = 2v and B' = B/2
∴
An electron of energy 1800 eV describes a circular path in magnetic field of flux density 0.4 T. The radius of path is (q = 1.6 x 10^{19} C, m_{e }= 9.1 x 10^{31} kg)
Two αpartides have the ratio of their velocities as 3 : 2 on entering the field. If they move in different circular paths, then the ratio of the radii of their paths is
As qvB = mv^{2}/r
When a positively charged particle enters a uniform magnetic field with uniform velocity, its trajectory can be (i) a straight line (ii) a circle (iii) a helix.
Two particles of equal charges after being accelerated through the same potential difference enter in a uniform transverse magnetic field and describe circular paths of radii R_{1 }and R_{2}. Then the ratio of their respective masses (M_{1}/M_{2}) is
And Bqv = Mv^{2}/R Or Using (i)
or
∴ M ∝ R^{2} (∵ B, q and V are same for the given two particles)
Hence (M_{1}/M_{2}) = (R_{1}/R_{2})^{2}
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