$\mathrm{L}=(3 \mathrm{a}) \mathrm{N} \quad(\mathrm{N}=\text { total tums })$
And length of winding $=(d) N=\ell$
$(d=\text { diameter of wire })$
Self inductance $=\mu_{0} \mathrm{n}^{2} \mathrm{A} \ell$
$=\mu_{0} n^{2}\left(\frac{\sqrt{3} a^{2}}{4}\right) d N$
$\alpha a^{2} N \propto a$
So self inductance will become $3$ times.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$1.$ 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 statement$(s)$ is(are)
$(A)$ 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 $V _2= V _1$
$2.$ Consider two different metallic strips ($1$ and $2$) of same dimensions (lengths $\ell$, 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 option$(s)$ is(are)
$(A)$ 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 $V_2=V_1$
Give the answer question $1$ and $2.$ 
