# 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 timesB)64 timesC)32 timesD)16 times

Samyak Jain
333 Points
4 years ago
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   $\dpi{100} \Rightarrow$  R = mv / qB   $\dpi{100} \Rightarrow$  v = qBR / m    ...(1)

Now, kinetic energy of proton, KE = (1/2)mv2  =  (1/2)m.q2B2R2 / 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.
Khimraj
3007 Points
4 years ago
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.