A fixed horizontal wire carries a current of | vedclass

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

Question

(C) For the second wire to float in air,the downward gravitational force must be balanced by the upward magnetic force exerted by the first wire.$1$.

Vedclass · Accepted Answer

$49\, A$ (direction of current is same to first wire)

Vedclass · Answer

(C) For the second wire to float in air,the downward gravitational force must be balanced by the upward magnetic force exerted by the first wire.$1$. The gravitational force per unit length is $F_g/l = (m/l)g = 10^{-2} 	imes 9.8 = 0.098\, N/m$.$2$. The magnetic force per unit length between two parallel wires is given by $F_m/l = \frac{\mu_0}{4\pi} \cdot \frac{2 i_1 i_2}{a}$.$3$. Equating the two forces: $(m/l)g = \frac{\mu_0}{4\pi} \cdot \frac{2 i_1 i_2}{a}$.$4$. Substituting the values: $10^{-2} 	imes 9.8 = 10^{-7} 	imes \frac{2 	imes 200 	imes i}{2 	imes 10^{-2}}$.$5$. Solving for $i$: $0.098 = 10^{-7} 	imes \frac{400 i}{0.02} = 10^{-7} 	imes 20000 i = 2 	imes 10^{-3} i$.$6$. $i = \frac{0.098}{2 	imes 10^{-3}} = 49\, A$.$7$. Since the magnetic force must be upward (repulsive) to counteract gravity,the currents in the two parallel wires must flow in the same direction.

Similar Questions

$A$ magnetic field is in the $+Y$ direction. $A$ current $i$ flows through the wire $PQRSTU$. Each side has a length $L$. What is the net force on the wire?

$A$ square current-carrying loop is suspended in a uniform magnetic field acting in the plane of the loop. If the force on one arm of the loop is $\overrightarrow{F}$,the net force on the remaining three arms of the loop is

Suppose an isolated north pole is kept at the centre of a circular loop carrying an electric current $i$. The magnetic field due to the north pole at a point on the periphery of the wire is $B$. The radius of the loop is $a$. The force on the wire is

$A$ 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