The magnetic field is directed normal to the plane of paper downwards. The proton emerges out of the magnetic field at point\[C\], then the distance \[AC\] and the value of angle \[\theta \] will respectively be
A) \[0.7m,\text{ }45{}^\circ \]
B) \[0.7\text{ }m,\,\,90{}^\circ \]
C) \[0.14\text{ }m,\text{ }90{}^\circ \]
D) \[0.14\text{ }m,\text{ }45{}^\circ \]
Correct Answer: D
Solution :
From the symmetry of figure, the angle \[\theta ={{45}^{o}}\]. The path of moving proton in a normal magnetic field is circular. If r is the radius of the circular path, then from the figure \[AC=2r\,\cos \,\,{{45}^{o}}=2r\times \frac{1}{\sqrt{2}}=\sqrt{2}r\] ?..(i) As \[Bqv=\frac{m{{v}^{2}}}{r}\] or \[r=\frac{mv}{Bq}\] \[AC=\frac{\sqrt{2}mv}{Bq}=\frac{\sqrt{2}\times 1.67\times {{10}^{-27}}\times {{10}^{7}}}{1\times 1.6\times {{10}^{-19}}}\] \[=0.14m\]
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