A charged particle is projected with velocity $\to v = u^i$ from the origin, in a region having electric field $\to E$ and magnetic field $\to B$ both. Value of $\to B$ and $\to E$ are given in column 1 and column 2 respectively and path of the particle is given in column 3.

$\begin{matrix} Column 1 & Column 2 & Column 3 (Magnetic field) & (Electric field) & (Path of particle) (i) & \to B = 0 & (i) & \to E = 0 & (P) & Circular path (i i) & \to B = B_{0}^i + B_{0}^j & (i i) & \to E = E_{0}^i & (Q) & Helical path (i i i) & \to B = B_{0}^j + B_{0}^k & (i i i) & \to E = B_{0} u (^j -^k) & (R) & Cycloid (i v) & \to B = B_{0}^k & (i v) & \to E = E_{0}^k & (S) & Straight line \end{matrix}$

In which of the following combination particle moves in uniform circular motion?

(IV) (i) (P)

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(III) (i) (P)

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(IV) (iv) (R)

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(IV) (iv) (Q)

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Solution

The correct option is A
(IV) (i) (P)

Since there is no electric field present and the magnetic field is perpendicular to the velocity of the particle, so the path of the particle will be uniform circular motion in this case.

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A charged particle is projected with velocity $\to v = u^i$ from the origin, in a region having electric field $\to E$ and magnetic field $\to B$ both. Value of $\to B$ and $\to E$ are given in column 1 and column 2 respectively and path of the particle is given in column 3.

$\begin{matrix} Column 1 & Column 2 & Column 3 (Magnetic field) & (Electric field) & (Path of particle) (i) & \to B = 0 & (i) & \to E = 0 & (P) & Circular path (i i) & \to B = B_{0}^i + B_{0}^j & (i i) & \to E = E_{0}^i & (Q) & Helical path (i i i) & \to B = B_{0}^j + B_{0}^k & (i i i) & \to E = B_{0} u (^j -^k) & (R) & Cycloid (i v) & \to B = B_{0}^k & (i v) & \to E = E_{0}^k & (S) & Straight line \end{matrix}$

In which of the following combination particle moves with uniform speed?

Which of the following combination is correct?

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 $\to v = 2 \frac{E_{0}}{B_{0}}^x$	$(i)$ $\to E = E_{0}^z$	$(P)$ $\to B = - B_{0}^x$
$(I I)$ 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?

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The correct option is A (IV) (i) (P) Since there is no electric field present and the magnetic field is perpendicular to the velocity of the particle, so the path of the particle will be uniform circular motion in this case.

The correct option is A
(IV) (i) (P)

Since there is no electric field present and the magnetic field is perpendicular to the velocity of the particle, so the path of the particle will be uniform circular motion in this case.