A straight wire of mass 0.2 ,kg and length 1. | vedclass

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

Question

(B) Given: Mass of the wire, $m = 0.2 \,kg$. Length of the wire, $l = 1.5 \,m$. Current, $I = 2 \,A$. Acceleration due to gravity, $g = 10 \,m/s^2$.Th

Vedclass · Accepted Answer

$0.67$

Vedclass · Answer

(B) Given: Mass of the wire, $m = 0.2 \,kg$. Length of the wire, $l = 1.5 \,m$. Current, $I = 2 \,A$. Acceleration due to gravity, $g = 10 \,m/s^2$.The magnetic force $F_m$ acting on a current-carrying wire in a uniform magnetic field is given by $F_m = I l B \sin 	heta$. Since the magnetic field is perpendicular to the wire, $	heta = 90^{\circ}$, so $F_m = I l B$.For the wire to be suspended in mid-air, the upward magnetic force must balance the downward gravitational force (weight) of the wire.$F_m = w$$I l B = m g$Rearranging to solve for $B$:$B = \frac{m g}{I l}$$B = \frac{0.2 	imes 10}{2 	imes 1.5}$$B = \frac{2}{3} \approx 0.67 \,T$Therefore, the magnitude of the magnetic field is $0.67 \,T$.

Similar Questions

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

$A, B$ and $C$ are three parallel conductors of equal lengths and carry currents $I, I$ and $2I$ respectively as shown in the figure. The distance between $A$ and $B$ and between $B$ and $C$ is $d$. If $F_1$ is the force exerted by $B$ on $A$ and $F_2$ is the force exerted by $C$ on $A$,then:

$A$ circular current loop of radius $a$ is placed in a radial magnetic field $B$ as shown in the figure. The net force acting on the loop is

$A$ rectangular loop of wire,supporting a mass $m$,hangs with one end in a uniform magnetic field $\vec B$ pointing into the plane of the paper. $A$ clockwise current $i$ is set up such that $i > mg/Ba$,where $a$ is the width of the loop. Then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module