A coil of Cu wire (radius r, self-inductance L ) is bent in two concentric turns each having radius (r/2). The self-inductance is now

Question

A coil of Cu wire (radius r, self-inductance L ) is bent in two concentric turns each having radius (r/2). The self-inductance is now (A) 2 L (B) L (C) 4L (D)  (L/2)

Tardigrade Admin · Accepted Answer

As L propto N2 r (L1/L2)=[(N1/N2)]2 × (r1/r2) ⇒ (L/L2)=[(1/2)]2 ×[(r/r / 2)]=(1/2) L12=2 L

Q. A coil of $C u$ wire (radius $r$ , self-inductance $L$ ) is bent in two concentric turns each having radius $\frac{r}{2}$ . The self-inductance is now

$2 L$

$L$

$4 L$

$\frac{L}{2}$

As $L \propto N^{2} r$
$\frac{L _{1}}{L _{2}} = [\frac{N _{1}}{N _{2}}]^{2} \times \frac{r _{1}}{r _{2}}$
$\Rightarrow \frac{L}{L _{2}} = [\frac{1}{2}]^{2} \times [\frac{r}{r /2}] = \frac{1}{2}$
$L_{12} = 2 L$

Q. A coil of Cu wire (radius r, self-inductance L ) is bent in two concentric turns each having radius 2r​. The self-inductance is now

2L

L

4L

2L​

As L∝N2r L2​L1​​=[N2​N1​​]2×r2​r1​​ ⇒L2​L​=[21​]2×[r/2r​]=21​ L12​=2L

Q. A coil of $C u$ wire (radius $r$ , self-inductance $L$ ) is bent in two concentric turns each having radius $\frac{r}{2}$ . The self-inductance is now

$2 L$

$L$

$4 L$

$\frac{L}{2}$

As $L \propto N^{2} r$
$\frac{L _{1}}{L _{2}} = [\frac{N _{1}}{N _{2}}]^{2} \times \frac{r _{1}}{r _{2}}$
$\Rightarrow \frac{L}{L _{2}} = [\frac{1}{2}]^{2} \times [\frac{r}{r /2}] = \frac{1}{2}$
$L_{12} = 2 L$