Question

a cyclotron adjusted to give proton beam magnetic induction is 0.15wb^-2and the extreme radius is 1.5m the energy of the emergent proton in MeV will be​

Answer 1

Answer:

★ KINETIC ENERGY ( KE ) OF PROTON BEAM = 2.5 MeV

Explanation:

★ CYCLOTRON ★

★ GIVEN ;

» Charge of Proton beam ( Q ) = 1.6 × 10^-19

» Magnetic Field ( B ) = 0.15 Weber / m²

» Extreme Radius ( R max ) = 1.5 m

» Mass of PROTON ( M ) = 1.6 × 10^-27 Kg

★ MAXIMUM KINETIC ENERGY GAINED BY CHARGE ( KE ) = ???

______ [ BY USING FORMULA ] ______

[$kinetic \: Energy \: ( \: ke \: ) \: = \: ( \frac{ {Q}^{2} \: \times \: {B}^{2} }{2 \: \times \: m} ) \: {Rmax}^{2} ]$

★ KE = [ ( 1.6 × 10^-19 )² × ( 0.15 )² / 2 × 1.6 × 10^-27 ] × ( 1.5 )²

» KE = [ 2.56 × 10^-38 × 225 × 10-⁴ / 3.2 × 10^-27 ] × 2.25

» KE = [ 256 × 10^-40 × 225 × 10-⁴ / 32 × 10^-28 ] × 225 × 10-²

» KE = [ 57,600 × 10-⁴⁴ / 32 × 10^-28 ] × 225 × 10-²

» KE = [ 1800 × 10^-16 ] × 225 × 10-²

» KE = [ 18 × 225 × 10^-16 ]

» KE = 4050 × 10^-16

★ » KINETIC ENERGY ( KE ) = 0.4 × 10-¹² Joules « ★

–––––––––– [ OR ] ––––––––––

★ » KINETIC ENERGY ( KE ) = 2.5 Mev [ Mega Electron Volt ] « ★

-------------- [ # conversions #] ---------------

» 1 Electron Volt ( ev ) = 1.6 × 10^-19 J

» Kinetic Energy ( KE ) = 4050 × 10^-16 J

[ " Kinetic Energy => Electron Volt " ]

4050 × 10³ × 10^-19 ( Power balance )

( 4050 / 1.6 ) × 10³ × 10^-19

25,31,250 × 10^-19 [ MEGA = 10^6 ]

★ 2.5 MeV ★

_________________________________________

ANSWER :- KINETIC ENERGY GAINED BY CHARGE HAVING MAGNETIC FIELD OF 0.15 W / m² is ;

★ KE = 2.5 Mega Electron Volt ★

Answer 2

Answer:

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