A) 4r
B) 2r
C) \[\frac{r}{2}\]
D) \[\frac{r}{4}\]
Correct Answer: A
Solution :
Mass of the electron = m Charge on the electron = q Initial radius of the circular path =5 Initial velocity of electron \[{{\upsilon }_{1}}=\upsilon \] Final velocity of the electron \[{{\upsilon }_{2}}=2\upsilon \] Angle which the direction of motion of the electron makes with the magnetic field \[\theta ={{90}^{o}}\] Initial magnetic field strength \[{{B}_{1}}=B\] Final magnetic field strength \[{{B}_{2}}=\frac{B}{2}\] The radius of the circular path in a uniform magnetic field is given by \[r=\frac{m\upsilon \sin \theta }{qB}=\frac{m\upsilon \sin {{90}^{o}}}{qB}=\frac{m\upsilon }{qB}\] \[\propto \frac{\upsilon }{B}\] Hence, \[\frac{{{r}_{1}}}{{{r}_{2}}}\times \frac{{{B}_{2}}}{{{B}_{1}}}=\frac{\upsilon }{2\upsilon }\times \frac{B/2}{B}\] \[{{r}_{2}}=4{{r}_{1}}=4r\]
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