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Grade 12Thermal Physics

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)1.6810^-6N (2)1.610^-10 N (3)0 (4) 3.2*10^-8

jayprakash pal

8 Years agoGrade 12

1 Answer

Eshan

8 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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