For a positively charged particle moving in an $x - y$ plane initially along the $x$ -axis, there is a sudden change in its path due to the presence of electric and/or magnetic fields beyond P. The curved path is shown in the $x - y$ plane and is found to be non-circular. Which one of the following combinations is possible [2003]

Question

Smart Achievers · Accepted Answer

(2)

When $\vec{E} = 0$ , then the path of the charged particle beyond P will be a helix; hence options (1) and (3) are incorrect. The velocity at P is in the $X$ -direction (given).

Let $\vec{v} = k \hat{i}$

After P, the positively charged particle gets deflected in the $x - y$ plane towards $- y$ direction and the path is non-circular.

$Now, \vec{F} = q (\vec{v} \times \vec{B})$

$\Rightarrow \vec{F} = q [k \hat{i} \times (c \hat{k} + a \hat{i})] for option (2)$

$= q [k c \hat{i} \times \hat{k} + k a \hat{i} \times \hat{i}]$ $= k c q (- \hat{j})$

Since in option (2), electric field is also present $\vec{E} = a \hat{i},$ therefore it will also exert a force in the $+ X$ direction. The net result of the two forces will be a non-circular path.

Only option (2) fits for the above logic.

Solution

1 $\vec{E} = 0; \vec{B} = b \hat{i} + c \hat{k}$

2 $\vec{E} = a \hat{i}; \vec{B} = c \hat{k} + a \hat{i}$

3 $\vec{E} = 0; \vec{B} = c \hat{j} + b \hat{k}$

4 $\vec{E} = a \hat{i}; \vec{B} = c \hat{k} + b \hat{j}$

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Solution

1 E→=0; B→=bi^+ck^

2 E→=ai^; B→=ck^+ai^

3 E→=0; B→=cj^+bk^

4 E→=ai^; B→=ck^+bj^

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1 $\vec{E} = 0; \vec{B} = b \hat{i} + c \hat{k}$

2 $\vec{E} = a \hat{i}; \vec{B} = c \hat{k} + a \hat{i}$

3 $\vec{E} = 0; \vec{B} = c \hat{j} + b \hat{k}$

4 $\vec{E} = a \hat{i}; \vec{B} = c \hat{k} + b \hat{j}$