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A proton is moving in a perpendicular magnetic field possess energy E .The magnetic fiepd is increased 8 times but proton is constrained to move in same radius. The KE will increase A)1/8 times B)64 times C)32 times D)16 times


one year ago

Samyak Jain
333 Points
							Ans. is B) 64 times.Solution :When a charge q is moving with a speed v in a perpendicular magnetic field B, it performs uniform circular motion.Let the radius of the circle described be R.We know that force due to magnetic field is F = q(v x B)Here velocity of proton is perpendicular to magnetic field. So magnitude of force isF = qvBThe centripetal force acting on charge is the force due to magnetic field on the charge.i.e.  mv2 / R  =  qvB   $\dpi{100} \Rightarrow$  R = mv / qB   $\dpi{100} \Rightarrow$  v = qBR / m    ...(1) Now, kinetic energy of proton, KE = (1/2)mv2  =  (1/2)m.q2B2R2 / m2i.e.  KE = q2B2R2 / 2m . As charge on proton q, its mass m are constant and radius R is same in both cases,KE $\alpha$  B2Thus, if B is increased 8 times, KE will increase 82 = 64 times.

one year ago
Khimraj
3007 Points
							kinetic energy of proton, KE = (1/2)mv2  =  (1/2)m.q2B2R2 / m2if B is increased 8 times, KE will increase 82 = 64 times.

one year ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions