Magnetics Ques 120
- Consider two different metallic strips ($1$ and $2$) of same dimensions (length $l$, width $w$ and thickness $d$ ) with carrier densities $n _1$ and $n _2$, respectively. Strip $1$ is placed in magnetic field $B _1$ and strip $2$ is placed in magnetic field $B _2$, both along positive $y$-directions. Then $V _1$ and $V _2$ are the potential differences developed between $K$ and $M$ in strips $1$ and $2$, respectively. Assuming that the current $I$ is the same for both the strips, the correct options is/are
(2015 Adv.)
(a) If $B _1=B _2$ and $n _1=2 n _2$, then $V _2=2 V _1$
(b) If $B _1=B _2$ and $n _1=2 n _2$, then $V _2=V _1$
(c) If $B _1=2 B _2$ and $n _1=n _2$, then $V _2=0.5 V _1$
(d) If $B _1=2 B _2$ and $n _1=n _2$, then $V _2=V _1$
Show Answer
Answer:
Correct Answer: 120.(a,c)
Solution:
- $V=\frac{B I}{n e d} \Rightarrow \frac{V _1}{V _2}=\frac{B _1}{B _2} \times \frac{n _2}{n _1}$
If $B _1=B _2$ and $n _1=2 n _2$, then $V _2=2 V _1$
If $B _1=2 B _2$ and $n _1=n _2$, then $V _2=0.5 V _1$
BETA
