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Grade: 12th pass
        
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
3 months ago

Answers : (2)

Samyak Jain
331 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 is
F = qvB
The centripetal force acting on charge is the force due to magnetic field on the charge.
i.e.  mv2 / R  =  qvB   \Rightarrow  R = mv / qB   \Rightarrow  v = qBR / m    ...(1)
 
Now, kinetic energy of proton, KE = (1/2)mv2  =  (1/2)m.q2B2R2 / m2
i.e.  KE = q2B2R2 / 2m . As charge on proton q, its mass m are constant and radius R is same in both cases,
KE \alpha  B2
Thus, if B is increased 8 times, KE will increase 82 = 64 times.
3 months ago
Khimraj
3008 Points
							
kinetic energy of proton, KE = (1/2)mv2  =  (1/2)m.q2B2R2 / m2
if B is increased 8 times, KE will increase 82 = 64 times.
3 months ago
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