The resultant force on the current loop PQRS | vedclass

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

Question

(D) The forces on segments $SR$ and $PQ$ are equal in magnitude but opposite in direction,so their net force is zero.For the segments $PS$ and $QR$ pa

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(D) The forces on segments $SR$ and $PQ$ are equal in magnitude but opposite in direction,so their net force is zero.For the segments $PS$ and $QR$ parallel to the long conductor,the force between two parallel conductors carrying currents $I_1$ and $I_2$ is given by:$F = \frac{\mu_0}{2\pi} \frac{I_1 I_2 \ell}{r} = 2 	imes 10^{-7} \frac{I_1 I_2 \ell}{r}$Here,$I_1 = 20 	ext{ A}$ (long conductor),$I_2 = 20 	ext{ A}$ (loop),and $\ell = 15 	ext{ cm} = 0.15 	ext{ m}$.For segment $PS$,$r_1 = 2 	ext{ cm} = 0.02 	ext{ m}$. The force is attractive (towards the conductor):$F_{PS} = \frac{2 	imes 10^{-7} 	imes 20 	imes 20 	imes 0.15}{0.02} = 6 	imes 10^{-4} 	ext{ N}$.For segment $QR$,$r_2 = 2 	ext{ cm} + 10 	ext{ cm} = 12 	ext{ cm} = 0.12 	ext{ m}$. The force is repulsive (away from the conductor):$F_{QR} = \frac{2 	imes 10^{-7} 	imes 20 	imes 20 	imes 0.15}{0.12} = 1 	imes 10^{-4} 	ext{ N}$.The net force is $F_{net} = F_{PS} - F_{QR} = 6 	imes 10^{-4} - 1 	imes 10^{-4} = 5 	imes 10^{-4} 	ext{ N}$.Thus,the value is $5$.

The resultant force on the current loop $PQRS$ due to a long current-carrying conductor will be $..... \times 10^{-4} \text{ N}$.

Similar Questions

$A$ triangular shaped wire carrying $10 A$ current is placed in a uniform magnetic field of $0.5 T$,as shown in the figure. The magnetic force on segment $CD$ is $.... N$ (Given $BC = CD = BD = 5 cm$).

$A$ 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 $\text{ g}$)

Two parallel very long straight wires carrying current of $5 \text{ A}$ each are kept at a separation of $1 \text{ m}$. If the currents are in the same direction,the force per unit length between them is . . . . . . $\text{N/m}$. $(\mu_0 = 4\pi \times 10^{-7} \text{ SI units})$

The magnetic force between the wires as shown in the figure is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module