Magnetics Ques 118
Passage Based Questions
In a thin rectangular metallic strip a constant current $I$ flows along the positive $x$-direction, as shown in the figure. The length, width and thickness of the strip are $l, w$ and $d$, respectively. A uniform magnetic field $\mathbf{B}$ is applied on the strip along the positive $y$-direction. Due to this, the charge carriers experience a net deflection along the $z$-direction.
This results in accumulation of charge carriers on the surface $P Q R S$ and appearance of equal and opposite charges on the face opposite to $P Q R S$. A potential difference along the $z$-direction is thus developed. Charge accumulation continues until the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross section of the strip and carried by electrons.
- Consider two different metallic strips ( $1$ and $2$) of the same material. Their lengths are the same, widths are $w _1$ and $w _2$ and thicknesses are $d _1$ and $d _2$, respectively. Two points $K$ and $M$ are symmetrically located on the opposite faces parallel to the $x-y$ plane (see figure). $V _1$ and $V _2$ are the potential differences between $K$ and $M$ in strips $1$ and $2$, respectively. Then, for a given current $I$ flowing through them in a given magnetic field strength $B$, the correct statements is/are
(2015 Adv.)
(a) If $w _1=w _2$ and $d _1=2 d _2$, then $V _2=2 V _1$
(b) If $w _1=w _2$ and $d _1=2 d _2$, then $V _2=V _1$
(c) If $w _1=2 w _2$ and $d _1=d _2$, then $V _2=2 V _1$
(d) If $w _1=2 w _2$ and $d _1=d _2$, then $V _2=V _1$
Show Answer
Answer:
Correct Answer: 118.(a,d)
Solution:
- $F _B=B e v=B e \frac{I}{n A e}=\frac{B I}{n A}$
$ F _e=e E $
$ \begin{aligned} F _e & =F _B \\ e E & =\frac{B I}{n A} \Rightarrow E=\frac{B}{n A e} \\ V & =E d=\frac{B I}{n A e} \cdot w=\frac{B I w}{n(w d) e}=\frac{B I}{n e d} \\ \frac{V _1}{V _2} & =\frac{d _2}{d _1} \\ \Rightarrow \text { if } \quad w _1 & =2 w _2 \end{aligned} $
and $d _1=d _2$
$ V _1=V _2 $
BETA
