A single-turn current loop in the shape of a | vedclass

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

Question

(B) The magnetic force on a current-carrying wire is given by $\vec{F} = I(\vec{L} 	imes \vec{B})$, where $L$ is the length vector of the wire.Let th

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(B) The magnetic force on a current-carrying wire is given by $\vec{F} = I(\vec{L} 	imes \vec{B})$, where $L$ is the length vector of the wire.Let the sides of the triangle be $EF = 5 \,cm$, $EG = 12 \,cm$, and $GF = 13 \,cm$.The magnetic field $\vec{B}$ is parallel to the side $GF$. Let $	heta$ be the angle between the side $EF$ and the side $GF$.From the geometry of the triangle, $\sin 	heta = \frac{EG}{GF} = \frac{12}{13}$.The force on the side $EF$ is $F_{EF} = I \cdot L_{EF} \cdot B \sin 	heta$.Given $I = 2 \,A$, $L_{EF} = 0.05 \,m$, $B = 0.75 \,T$, and $\sin 	heta = \frac{12}{13}$.$F_{EF} = 2 	imes 0.05 	imes 0.75 	imes \frac{12}{13} = 0.1 	imes 0.75 	imes \frac{12}{13} = 0.075 	imes \frac{12}{13} = \frac{0.9}{13} = \frac{9}{130} \,N$.Comparing this with $\frac{x}{130} \,N$, we get $x = 9$.

Similar Questions

In the given diagram, the loop tends to:

$A$ conducting bar $PQ$ of length $l$ carrying current $I$ is suspended from a rigid support as shown in the figure. $A$ uniform magnetic field $B$ perpendicular to $PQ$ and directed away from the reader (into the plane) is applied. If the mass of the bar is $M$,find the tension in each string.

$A$ conducting circular loop of radius $r$ carries a constant current $i$. It is placed in a uniform magnetic field $B$,such that $B$ is perpendicular to the plane of the loop. The net magnetic force acting on the loop is

What is the net force on the rectangular coil?

Vedclass Test Series

Exam Paper Generator

Online Exam Module