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AMU Medical AMU Solved Paper-2004

  • question_answer
    A beam of ions enters normally into a uniform magnetic field of \[\text{4}\times \text{1}{{0}^{-2}}\] T with velocity \[\text{2}\times \text{1}{{0}^{\text{5}}}\text{ m}/\text{s}.\]If the specific charge of the ion - \[\text{5}\times \text{l}{{0}^{\text{7}}}\text{ C}/\text{kg},\]Then the radius of the circular path described will be

    A)  0.10 m                                 

    B)  0.06 m

    C)  0.20 m                                 

    D)  0.25 m

    Correct Answer: A

    Solution :

                     Force due to magnetic field provides the required centripetal force to the ions. \[\therefore \]Centripetal force = force due to magnetic field                 \[\frac{m{{v}^{2}}}{r}=evB\,\sin \theta \] Given,   \[\theta ={{90}^{o}},\frac{e}{m}=5\times {{10}^{7}}\,C/kg\],                 \[B=4\times {{10}^{-2}}T\],                 \[v=2\times {{10}^{5}}m/s\] \[\therefore \]  \[r=\frac{mv}{eB}\] \[\Rightarrow \]               \[r=\frac{2\times {{10}^{5}}}{5\times {{10}^{7}}\times 4\times {{10}^{-2}}}=0.1\,m\]      


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