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.
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