A current-carrying rectangular loop PQRS is m | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The magnetic force per unit length $f$ between two parallel wires carrying currents $I_1$ and $I_2$ separated by a distance $d$ is given by the fo

Vedclass · Accepted Answer

$1: 4$

Vedclass · Answer

(B) The magnetic force per unit length $f$ between two parallel wires carrying currents $I_1$ and $I_2$ separated by a distance $d$ is given by the formula:$f = \frac{\mu_0 I_1 I_2}{2\pi d}$In this rectangular loop,the current $I$ flows through both wires $PQ$ and $RS$. Therefore,$I_1 = I_2 = I$ (or $2I$ in the second case).Thus,the force per unit length is proportional to the square of the current: $f \propto I^2$.For the first case with current $I$,$f_{PQ}^{I} \propto I^2$.For the second case with current $2I$,$f_{PQ}^{2I} \propto (2I)^2 = 4I^2$.Taking the ratio:$\frac{f_{PQ}^{I}}{f_{PQ}^{2I}} = \frac{I^2}{4I^2} = \frac{1}{4}$.Therefore,the ratio is $1: 4$.

Similar Questions

$A$ long current-carrying conductor is placed near a rectangular current loop as shown in the figure. Calculate the resultant force on the current loop.

In the given figure,what is the ratio of the magnetic force on wire $ab$ to that on wire $bc$? (Given $ab = l$ and $\angle abc = 45^o$)

Three long,straight parallel wires carrying currents are arranged as shown. The wire $C$ which carries a current of $5.0 \text{ A}$ is so placed that it experiences no force. The distance of wire $C$ from wire $D$ is (in $cm$)

$A$ straight wire of length $50 \text{ cm}$ carrying a current of $2.5 \text{ A}$ is suspended in mid-air by a uniform magnetic field of $0.5 \text{ T}$. Calculate the mass of the wire. (Take $g = 10 \text{ m s}^{-2}$) (in $\text{ g}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module