By Sunil Bhardwaj

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In cyclotron, the magnetic field is applied to move particle in semicircular path. If H is the strength of magnetic field and e is the charge possessed by the particle and v is the velocity of the particle, then the magnetic force acting on the particle is Hev and centrifugal force is $$\frac { m{ v }^{ 2 } }{ r } .$$ on equating the two. $$ Hev = \frac { m{ v }^{ 2 } }{ r } $$ $$ \therefore Her = mv $$ or $$v = \frac { Her }{ m } \qquad \qquad \qquad \qquad \qquad ...(1) $$ The distance covered by any in any dee apparatus is half of the circumference i.e. \(\pi r\) then, $$ Velocity = \frac { Distance }{ Time } $$ or $$Time (t) = \frac { Distance }{ Velocity } $$ $$ \therefore t = \frac { \pi r }{ v } $$ lets put the value of v $$ t = \frac { \pi r }{ \left( \frac { Her }{ m } \right) } = \frac { \pi m }{ He } \qquad \qquad \qquad ...(2) $$ In eq (2) there is no v and r so the time taken by the projectile is independent of v and r and is constant. The energy possessed by the projectiles is only KE, $$ KE = \frac { 1 }{ 2 } m{ v }^{ 2 } $$ On putting value of v from eq (1) $$ KE = \frac { 1 }{ 2 } m{ \left( \frac { Her }{ m } \right) }^{ 2 } $$ $$ KE = \frac { 1 }{ 2 } \frac { { \left( Her \right) }^{ 2 } }{ m } $$ $$ \therefore \boxed { { E }_{ max } = \frac { { H }^{ 2 }{ e }^{ 2 }{ r }_{ max }^{ 2 } }{ 2m } } $$ Therefore to increase E, we must increase r i.e. radius of curvature because all the other factors are constant.