One proton beam enters in a magnetic field of 10-4 T normally with specific charge 1011 C/kg and velocity of 107 m/s. What will be the radius of circle?
Correct answer: B. 1 m
- A. 0.1 m
- B. 1 m
- C. 10 m
- D. none of them
Explanation
The radius of proton in the circular path is given by , r= mv/ qB Where , V = 10^7 q= 10^11B = 10^-4 Therefore the equation becomes , r = (10^7) / { ( 10^ 11 ) x ( 10^-4 ) } Radius of circle = 1 m ∵q/m = 10^11 C/kg Thus only option B is correct
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About Electromagnetism
Magnetic flux density and magnetic flux describe the strength of a magnetic field and the field passing through a surface. A charged particle moving through a magnetic field experiences a force perpendicular to its velocity and may follow circular or helical motion, depending on the angle between velocity and field.
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