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Moving Charges and Magnetism question

2012 · Q134
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Moving Charges and Magnetism question

2012 · Q134

NEETPhysicsMoving Charges and MagnetismMCQ+4 / −1
An alternating electric field, of frequency vvv, is applied across the does (radius = R) of a cyclotron that is being used to accelerate protons (mass = m). The operating magnetic field (B) used in the cyclotron and the kinetic energy (K) of the proton beam, produced by it, are given by
  1. A
    B=mυeB = {{m\upsilon } \over e}B=emυ​  and  K=2mπ2υ2R2K = 2m{\pi ^2}{\upsilon ^2}{R^2}K=2mπ2υ2R2
  2. B
    B=2πmυeB = {{2\pi m\upsilon } \over e}B=e2πmυ​  K=m2πυR2K = {m^2}\pi \upsilon {R^2}K=m2πυR2
  3. C
    B=2πmυeB = {{2\pi m\upsilon } \over e}B=e2πmυ​  K=2mπ2v2R2K = 2m{\pi ^2}{v^2}{R^2}K=2mπ2v2R2
  4. D
    B=mυeB = {{m\upsilon } \over e}B=emυ​  K=m2πυR2K = {m^2}\pi \upsilon {R^2}K=m2πυR2
View written solutionFree

Correct answer: C

Time period of cyclotron is

T=1v=2πmeB;B=2πmev;R=mveB=peBT = {1 \over v} = {{2\pi m} \over {eB}}; B = {{2\pi m} \over e}v;R = {{mv} \over {eB}} = {p \over {eB}}T=v1​=eB2πm​;B=e2πm​v;R=eBmv​=eBp​

⇒P=eBR=e×2πmveR=2πmvR \Rightarrow P = eBR = e \times {{2\pi mv} \over e}R = 2\pi mvR⇒P=eBR=e×e2πmv​R=2πmvR

K.E.=p22m=(2πmvR)22m=2π2mv2R2K.E. = {{{p^2}} \over {2m}} = {{{{\left( {2\pi mvR} \right)}^2}} \over {2m}} = 2{\pi ^2}m{v^2}{R^2}K.E.=2mp2​=2m(2πmvR)2​=2π2mv2R2

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