KVPY Sample Paper KVPY Stream-SX Model Paper-5

  • 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 :

    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 \[{{P}_{f}}=mv\hat{i}+mv\theta \hat{j}\]
    \[\Delta P=mv\theta \hat{j}\]\[=mv\times \frac{d}{r}=\frac{mv\times d\times qB}{mv}\]
    Impulse =qBd   

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