Magnetics 1 Question 24
24. A proton moving with a constant velocity passes through a region of space without any change in its velocity. If $E$ and $B$ represent the electric and magnetic fields respectively. Then, this region of space may have
$(1985,2 M)$
(a) $E=0, B=0$
(b) $E=0, B \neq 0$
(c) $E \neq 0, B=0$
(d) $E \neq 0, B \neq 0$
Match the Columns
Directions (Q.Nos. 25-27) Matching the information given in the three columns of the following table.
A charged particle (electron or proton) is introduced at the origin $(x=0, y=0, z=0)$ with a given initial velocity $\mathbf{v}$. A uniform electric field $\mathbf{E}$ and a uniform magnetic field $\mathbf{B}$ exist everywhere. The velocity $\mathbf{v}$, electric field $\mathbf{E}$ and magnetic field $\mathbf{B}$ are given in columns 1, 2 and 3, respectively. The quantities $E _0, B _0$ are positive in magnitude.
| Column 1 | Column 2 | Column 3 | ||
|---|---|---|---|---|
| (l) | Electron with $v=2 \frac{E _0}{B _0} \hat{x}$ |
(i) | $E=E _0 \hat{z}$ | (P) $B=-B _0 \hat{x}$ |
| (II) | Election with $v=\frac{E _0}{B _0} \hat{y}$ |
(ii) | $E=-E _0 \hat{y}$ | (Q) $B=B _0 \hat{x}$ |
| (III) | Proton with $v=0$ |
(iii) | $E=-E _0 \hat{x}$ | (R) $B=B _0 \hat{y}$ |
| (IV) | Proton with $v=2 \frac{E _0}{B _0} \hat{x}$ |
(iv) | $E=E _0 \hat{x}$ | (S) $B=B _0 \hat{z}$ |
(2017 Adv.)
Show Answer
Answer:
Correct Answer: 24. $(a, b, d)$
Solution:
- If both $E$ and $B$ are zero, then $\mathbf{F} _e$ and $\mathbf{F} _m$ both are zero. Hence, velocity may remain constant. Therefore, option (a) is correct.
If $E=0, B \neq 0$ but velocity is parallel or antiparallel to magnetic field, then also $\mathbf{F} _e$ and $\mathbf{F} _m$ both are zero. Hence, option (b) is also correct.
If $E \neq 0, B \neq 0$ but $\mathbf{F} _e+\mathbf{F} _m=0$, then also velocity may remain constant or option (d) is also correct.
