A charged particle (electron or proton) is introduced at the origin $(x = 0, y = 0, x = 0)$ with a given initial velocity $\to v$ . A uniform electric field $\to E$ and a uniform magnetic field $\to B$ exist everywhere. The velocity $\to v$ , electric field $\to E$ and magnetic field $\to B$ are given in column $1, 2$ and $3,$ respectively. The quantities $E_{0}, B_{0}$ are positive in magnitude.

Column 1

Column 2

Column 3

(I) Electron with $\to v = 2 \frac{E_{0}}{B_{0}}^x$

(i) $\to E = E_{0}^z$

(P) $\to B = - B_{0}^x$

(II) Electron with $\to v = \frac{E_{0}}{B_{0}}^y$

(ii) $\to E = - E_{0}^y$

(Q) $\to B = - B_{0}^x$

(III) Electron with $\to v = 0$

(iii) $\to E = - E_{0}^x$

(R) $\to B = B_{0}^y$

(IV) Electron with $\to v = 2 \frac{E_{0}}{B_{0}}^x$

(iv) $\to E = E_{0}^x$

(S) $\to B = B_{0}^z$

In which case will the particle describe a helical path with axis along the positive $z$ direction?

(II) (ii) (R)

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(IV)(i)(S)

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(IV)(ii)(R)

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(III) (iii) (P)

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Solution

The correct option is B (IV)(i)(S)
In a given situation, a system will obtain the helical path when magnetic field $\to B$ and electric field $\to E$ direction are same (in $^z$ direction).

For A, C and D:
Force due to $\to B$ is zero but particle will accelerate due to $\to E$

For option-B:
$\to V ⊥ \to E ⊥ \to B$ and $v = \frac{| E |}{| B |}$
Hence, it will travel in straight line so net flux on it is zero.
Hence, option $B$ is correct.

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Similar questions

A charged particle (electron or proton) is introduced at the origin $(x = 0, y = 0, x = 0)$ with a given initial velocity $\to v$ . A uniform electric field $\to E$ and a uniform magnetic field $\to B$ exist everywhere. The velocity $\to v$ , electric field $\to E$ and magnetic field $\to B$ are given in column $1, 2$ and $3,$ respectively. The quantities $E_{0}, B_{0}$ are positive in magnitude.

Column 1	Column 2	Column 3
$(I)$ Electron with $\to v = 2 \frac{E_{0}}{B_{0}}^x$	$(i)$ $\to E = E_{0}^z$	$(P)$ $\to B = - B_{0}^x$
$(I I)$ Electron with $\to v = \frac{E_{0}}{B_{0}}^y$	$(i i)$ $\to E = - E_{0}^y$	$(Q)$ $\to B = - B_{0}^x$
$(I I I)$ Electron with $\to v = 0$	$(i i i)$ $\to E = - E_{0}^x$	$(R)$ $\to B = B_{0}^y$
$(I V)$ Electron with $\to v = 2 \frac{E_{0}}{B_{0}}^x$	$(i v)$ $\to E = E_{0}^x$	$(S)$ $\to B = B_{0}^z$

In which case will the particle move in a straight line with constant velocity?

Column 1	Column 2	Column 3
(I) Electron with $\to v = 2 \frac{E_{0}}{B_{0}}^x$	(i) $\to E = E_{0}^z$	(P) $\to B = - B_{0}^x$
(II) Electron with $\to v = \frac{E_{0}}{B_{0}}^y$	(ii) $\to E = - E_{0}^y$	(Q) $\to B = - B_{0}^x$
(III) Electron with $\to v = 0$	(iii) $\to E = - E_{0}^x$	(R) $\to B = B_{0}^y$
(IV) Electron with $\to v = 2 \frac{E_{0}}{B_{0}}^x$	(iv) $\to E = E_{0}^x$	(S) $\to B = B_{0}^z$

In which case will the particle describe a helical path with axis along the positive

z

direction?

Column 1	Column 2	Column 3
$(I)$ Electron with $\to v = 2 \frac{E_{0}}{B_{0}}^x$	$(i)$ $\to E = E_{0}^z$	$(P)$ $\to B = - B_{0}^x$
$(I I)$ Electron with $\to v = \frac{E_{0}}{B_{0}}^y$	$(i i)$ $\to E = - E_{0}^y$	$(Q)$ $\to B = - B_{0}^x$
$(I I I)$ Electron with $\to v = 0$	$(i i i)$ $\to E = - E_{0}^x$	$(R)$ $\to B = B_{0}^y$
$(I V)$ Electron with $\to v = 2 \frac{E_{0}}{B_{0}}^x$	$(i v)$ $\to E = E_{0}^x$	$(S)$ $\to B = B_{0}^z$