question_answer

A charged particle of mass m and charge q is projected into a uniform magnetic field of induction B with speed v which is perpendicular to B. The width of the magnetic field is d. The impulse imparted to fee particle by the field is \[\left( d<<\frac{mv}{qB} \right)\]

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