Magnetic Effects & Magnetism

JEE Main level, single correct, medium

171 of 276

Question 171 of 276

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 RR, assuming the neutron and proton masses are nearly equal, the radius of the proton's path will be

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Answer: (B) R2\frac{R}{\sqrt{2}}

As charge on both proton and deuteron is same i.e. ' ee '
Energy acquired by both, E=eVE=\mathrm{eV} For Deuteron.
Kinetic energy, 12mV2=eV\frac{1}{2} m V^2=e V
[ VV is the potential difference]
v=2eVmdv=\sqrt{\frac{2 e V}{m_d}}
But md=2mm_d=2 m
Therefore, v=2eV2m=eVmv=\sqrt{\frac{2 e V}{2 m}}=\sqrt{\frac{e V}{m}}
Radius of path, R=mveBR=\frac{m v}{e B}
Substituting value of ' vv ' we get
R=2mevmeBR2=mevmeB\begin{array}{r} R=\frac{2 m \sqrt{\frac{e v}{m}}}{e B} \\ \frac{R}{2}=\frac{m \sqrt{\frac{e v}{m}}}{e B} \end{array}
For proton :
12mV2=eVV=2eVm Radius of path, R′=mVeB=m2eVmeBR′=2×R2 [From eq. (i)] R′=R2\begin{aligned} & \frac{1}{2} m V^2=e V \\ & V=\sqrt{\frac{2 e V}{m}} \\ & \text { Radius of path, } R^{\prime}=\frac{m V}{e B}=\frac{m \sqrt{\frac{2 e V}{m}}}{e B} \\ & R^{\prime}=\sqrt{2} \times \frac{R}{2} \quad \text { [From eq. (i)] } \\ & R^{\prime}=\frac{R}{\sqrt{2}} \end{aligned}

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