Two wires A and B of the same material- having radii in the ratio 1 - 2 and carry currents in the ratio 4 - 1- The ratio of drift speed of electrons in A and B is16 - 14 - 11 - 41 - 16

Question

Toppr Admin · Accepted Answer

Current flowing through the conductor- I-nevA-xA0-Hence41-nevd1-x3C0-1-2nevd2-2-2orvd1vd2-4-xD7-41-161Hence correct answer is option A-

Two wires $A$ and $B$ of the same material, having radii in the ratio $1 : 2$ and carry currents in the ratio $4 : 1$ . The ratio of drift speed of electrons in $A$ and $B$ is
$16 : 1$
$4 : 1$
$1 : 4$
$1 : 16$

Current flowing through the conductor, $I = n e v A$ .
Hence
$\frac{4}{1} = \frac{n e v_{d_{1}} π (1)^{2}}{n e v_{d_{2}} (2)^{2}}$
or

$\frac{v_{d_{1}}}{v_{d_{2}}} = \frac{4 \times 4}{1} = \frac{16}{1}$
Hence correct answer is option A.

Two wires A and B of the same material, having radii in the ratio 1:2 and carry currents in the ratio 4:1. The ratio of drift speed of electrons in A and B is16:14:11:41:16

Current flowing through the conductor, I=nevA. Hence41=nevd1π(1)2nevd2(2)2orvd1vd2=4×41=161Hence correct answer is option A.

Two wires $A$ and $B$ of the same material, having radii in the ratio $1 : 2$ and carry currents in the ratio $4 : 1$ . The ratio of drift speed of electrons in $A$ and $B$ is
$16 : 1$
$4 : 1$
$1 : 4$
$1 : 16$

Current flowing through the conductor, $I = n e v A$ .
Hence
$\frac{4}{1} = \frac{n e v_{d_{1}} π (1)^{2}}{n e v_{d_{2}} (2)^{2}}$
or

$\frac{v_{d_{1}}}{v_{d_{2}}} = \frac{4 \times 4}{1} = \frac{16}{1}$
Hence correct answer is option A.