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a Proton of 200 Me V energy enters the magnetic field of 5 T. if direction of field is from South to North and motion is upwards the force acting on it will be: (1)1.68*10^-6N (2)1.6*10^-10 N (3)0 (4) 3.2*10^-8

a Proton of 200 Me V  energy enters  the magnetic field of 5 T.  if direction of field is from South to North and motion is upwards the force acting on it will be:
(1)1.68*10^-6N    (2)1.6*10^-10 N   (3)0     (4) 3.2*10^-8 

Grade:12

1 Answers

Eshan
askIITians Faculty 2095 Points
5 years ago
Dear student,

Kinetic energy of proton=K=\dfrac{1}{2}mv^2
v=\sqrt{\dfrac{2K}{m}}
Hence the force acting on proton=qvBsin90^{\circ}
=e\sqrt{\dfrac{2K}{m}}B=1.6\times 10^{-10}N

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