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. \[=\frac{(2\times {{10}^{5}}m/s)\times 2\times 3.14\times 0.012m}{2\sqrt{3}\times {{10}^{5}}m/s}\] \[=0.044m=4.4cm\]You need to login to perform this action.
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