2. Solved Papers
  3. EAMCET Medical

EAMCET Medical EAMCET Medical Solved Paper-2009

  • question_answer
    A charged particle with velocity \[\overrightarrow{v}=x\overrightarrow{i}+y\widehat{j}\] moves in a magnetic field \[\overrightarrow{B}=y\overrightarrow{i}+x\widehat{j}.\] Magnitude of the force acting on the parade is F. The correct option for F is (i) No force will act on panicle if x = y (ii) Force will act along y-axis if y < x (iii) Force is proportional to \[\left( {{x}^{2}}-{{y}^{2}} \right)\]if x > y (iv) Force is proportional to (x2 + y2) if y > x

    A)  (i) and (ii) are true

    B)  (i) and (iii) are true

    C)  (ii) and (iv) are true

    D)  (iii) and (iv) are true

    Correct Answer: B

    Solution :

     By Lorentz force equation \[\vec{F}=q\vec{v}\times \vec{B}=q(x\hat{i}+y\hat{j})\times (y\hat{i}+x\hat{j})\] \[\vec{v}\times \vec{B}=\left| \begin{matrix}    {\hat{i}} & {\hat{j}} & {\hat{k}}  \\    x & y & 0  \\    y & x & 0  \\ \end{matrix} \right|\]                 \[=\hat{i}(0)-\hat{j}(0)+\hat{k}({{x}^{2}}-{{y}^{2}})\] \[\therefore \]  \[\vec{F}=q.\hat{k}({{x}^{2}}-{{y}^{2}})\] \[\therefore \]  \[|\vec{F}|\propto ({{x}^{2}}-{{y}^{2}})\] If \[x=y,\]then direction of motion of charge  particle is same as direction of magnetic field then no force is applied on the charge.


You need to login to perform this action.
You will be redirected in 3 sec spinner