A) \[B=\frac{mv}{qd}\]and\[\Delta t=\frac{\pi d}{v}\]
B) \[B=\frac{mv}{2qd}\]and\[\Delta t=\frac{\pi d}{2v}\]
C) \[B=\frac{2mv}{qd}\]and\[\Delta t=\frac{\pi d}{2v}\]
D) \[B=\frac{2mv}{qd}\]and\[\Delta t=\frac{\pi d}{v}\]
Correct Answer: C
Solution :
| [c] The applied magnetic field provides the required centripetal force to the charge particle, so it can move in circular path of radius\[\frac{d}{2}\] |
| \[\therefore \]\[Bqv=\frac{m{{v}^{2}}}{d/2}\]or\[B=\frac{2mv}{qd}\] |
| Time interval for which a uniform magnetic field is applied\[\Delta t=\frac{\pi .\frac{d}{2}}{v}\](particle reverses its direction after time \[\Delta t\]by covering semi-circle). \[\Delta t=\frac{\pi d}{2v}\] |
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