A charged particle moves in a gravity-free space without change in velocity. Which of the following is/are possible?

E = 0, B = 0

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E = 0, B \neq 0

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E \neq 0, B = 0

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E \neq 0, B \neq 0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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Solution

The correct options are
A $E = 0, B \neq 0$
B $E = 0, B = 0$
C $E \neq 0, B \neq 0$
If A charged particle moves in a gravity-free space without change in velocity, then
(1) Particle can move with constant velocity in any direction, so $B = 0, E = 0$
(2) Particle can move in a circle with constant speed. Magnetic force will provide the centripetal force that causes particle is move in a circle.
(3) If $q E = q v B$ and Magnetic & Electric force in opposite direction in this case also particle move with uniform speed.

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Similar questions

Q. A charged particle moves in a gravity-free space without change in velocity. Which of the following is/are possible?
(a) E = 0, B = 0
(b) E = 0, B ≠ 0
(c) E ≠ 0, B = 0
(d) E ≠ 0, B ≠ 0

Q. A charged particle moves along a circle under the action of possible constant electric and magnetic fields. Which of the following is possible?
(a) E = 0, B = 0
(b) E = 0, B ≠ 0
(c) E ≠ 0, B = 0
(d) E ≠ 0, B ≠ 0

Q. A proton moving with a constant speed passes through a region of space without any change in its speed. If

E

and

B

represent the electric and magnetic fields respectively, this region of space may have
a)

B \neq 0, E = 0

B = 0, E \neq 0

B = 0, E = 0

B \neq 0, E \neq 0

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 would the particle move in a straight line along the negative direction of

y

-axis (i.e, move along

-^y) ?

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