A proton and a deuteron are both accelerated through the same potential difference and enter in a magnetic field perpendicular to the direction of…
JEE Main 2012 — Physics Electromagnetism
2012mcqmedium
A proton and a deuteron are both accelerated through the same potential difference and enter in a magnetic field perpendicular to the direction of the field. If the deuteron follows a path of radius R, assuming the neutron and proton masses are nearly equal, the radius of the proton's path will be
Official previous-year question
Held on 19 May 2012 · Verified 6 Jul 2026.
Options
A
2R
B
2R
C
2R
D
R
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Solution
As charge on both proton and deuteron is same i.e. ' e ' Energy acquired by both, E=eV For Deuteron. Kinetic energy, 21mV2=eV [ V is the potential difference] v=md2eV But md=2m Therefore, v=2m2eV=meV Radius of path, R=eBmv Substituting value of ' v ' we get R=eB2mmev2R=eBmmev For proton : 21mV2=eVV=m2eV Radius of path, R′=eBmV=eBmm2eVR′=2×2R [From eq. (i)] R′=2R
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