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Moving Charges And Magnetism

Question
CBSEENPH12039940
Wired Faculty App

A proton and an alpha particle both enter a region of uniform magnetic field B, moving at angles to the field B. If the radius of circular orbits for both the particles is equal and the kinetic energy acquired by proton is 1 MeV, the energy acquired by the alpha particle will be

  • 4 MeV

  • 0.5 Mev

  • 1.5 MeV

  • 1 MeV

Solution

D.

1 MeV

Radius in magnetic field of circular orbit,
straight R space equals space mV over qB space equals space fraction numerator square root of 2 mE end root over denominator qB end fraction
and total energy of a moving particle in a circular orbit,straight E space equals fraction numerator space straight q squared straight B squared straight R squared over denominator 2 straight m end fraction space For space straight a space proton space enter space in space straight a space region space of space magnetic space field comma straight E subscript 1 space equals space fraction numerator straight e squared space straight x space straight B squared space straight x space straight R squared over denominator 2 space straight x space straight m subscript straight P end fraction space space space.... space left parenthesis straight i right parenthesis Where space straight m subscript straight P space is space the space mass space of space proton. Similarly comma space for space straight a space straight alpha space minus space particle space moves space in space straight a space uniform space magnetic space field straight E subscript 2 space equals space fraction numerator left parenthesis 2 straight e right parenthesis squared space straight x space straight B squared space straight x space straight R over denominator 2 space straight x space left parenthesis 4 straight m subscript straight P right parenthesis end fraction space left square bracket space straight m subscript straight alpha equals space 4 straight m subscript straight p right square bracket space.... space left parenthesis ii right parenthesis Dividing space Eq. space left parenthesis ii right parenthesis space by space Eq. space left parenthesis straight i right parenthesis space we space get straight E subscript 2 over straight E subscript 1 equals space fraction numerator left parenthesis 2 straight e right parenthesis squared space straight x space straight B squared space straight x space straight R over denominator 2 space straight x space left parenthesis 4 straight m subscript straight P right parenthesis end fraction space straight x space fraction numerator 2 space straight x space straight m subscript straight P over denominator straight e squared space straight x space straight B squared space straight x space straight R squared end fraction straight E subscript 2 over straight E subscript 1 space equals space 1 space rightwards double arrow space straight E subscript 2 space equals straight E subscript 1 space equals space 1 space MeV

Some More Questions From Moving Charges And Magnetism Chapter

Two moving coil meters, M1 and M2 have the following particulars:
R1 = 10 Ω   N1 = 30,
A1 = 3.6 x 10–3 m2,  B1 = 0.25 T
R2 = 14 Ω;  N2 = 42,
A2 = 1.8 x 10–3 m2, B2 = 0.50 T
(The spring constants k are identical for the two meters).
Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M2 and M1.

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