A) \[qBv\]
B) \[\frac{mv}{qB}\]
C) \[qBd\]
D) \[\frac{2\,m{{v}^{2}}}{qB}\]
Correct Answer: C
Solution :
[C]q more on circular path in B. |
Whose radius \[=\frac{mv}{qB}=r\] |
In question width of magnetic field region is very small than r. so charge q will exit from B on opposite side of region path of q is shown in figure. |
As \[d<\,<\,<\frac{mv}{qB}\] |
Using \[\sin \theta =\frac{d}{r}\] |
Impulse is change in momentum |
Where \[r=\frac{mv}{qB}\] |
\[\therefore d<<<r\] \[\therefore \sin \theta =\theta \] |
\[\theta =\frac{d}{r}\] |
Change in momentum \[\Delta P={{P}_{f}}-{{P}_{i}}\] |
Where \[{{P}_{f}}=mv\hat{i}+mv\theta \hat{j}\] |
\[{{P}_{i}}=mv\hat{i}\] |
\[\Delta P=mv\theta \hat{j}\]\[=mv\times \frac{d}{r}=\frac{mv\times d\times qB}{mv}\] |
Impulse =qBd |
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