• # question_answer A beam of protons with a velocity of $4\times {{10}^{5}}\,m/s$enters in a uniform magnetic field of$3.0T$. The velocity makes an angle of ${{60}^{o}}$ with the magnetic field. The radius of the helical path taken by the proton beam, and the pitch of the helix, are A) $1.2cm$and $4.4cm~~$ B)  $1.4cm$and $4.2cm$ C)  $2.4cm$and $3.1cm$ D)  None of these

Correct Answer: A

Solution :

The component of the proton's velocity along and perpendicular to the magnetic field are ${{V}_{||}}=(4\times {{10}^{5}}m/s)\,\sin {{60}^{o}}$ $=2\times {{10}^{5}}m/s$ and           ${{V}_{\bot }}=(4\times {{10}^{5}}m/s)\,\,sin{{60}^{o}}$                 $=2\sqrt{3}\times {{10}^{5}}m/s$ As the force $qV\times B$ is perpendicular to the magnetic field, the component ${{V}_{||}}$ will remain constant. In the plane perpendicular to the field, the proton will discribe a circle whose radius is obtained from the equation.                                 $q{{V}_{\bot }}B=\frac{mV_{\bot }^{2}}{r}$ or         $r=\frac{m{{V}_{||}}}{qB}$                 $=\frac{(1.67\times {{10}^{-27}}kg)\times (2\sqrt{3}\times {{10}^{5}}m/s)}{(1.6\times {{10}^{-19}}C)\times (0.3T)}$                 $=0.012=1.2cm$ The time taken in one complete revolution in the plane perpendicular to B is.                                 $T=\frac{2\pi r}{{{V}_{\bot }}}=\frac{2\times 3.14\times 0.012m}{2\sqrt{3}\times {{10}^{5}}m/s}$ The distance moved along the field during this period i.e. the pitch.             $=\frac{(2\times {{10}^{5}}m/s)\times 2\times 3.14\times 0.012m}{2\sqrt{3}\times {{10}^{5}}m/s}$                 $=0.044m=4.4cm$

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