A long straight wire carrying a current of 25 | vedclass

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

Question

(C) The magnetic force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated by a distance $h$ is given by $F/l = \fr

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) The magnetic force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated by a distance $h$ is given by $F/l = \frac{\mu_0 I_1 I_2}{2 \pi h}$.Since the currents are in opposite directions,the force is repulsive.At equilibrium,the repulsive magnetic force balances the gravitational force acting on wire $PQ$:$F = mg$$\frac{\mu_0 I^2 l}{2 \pi h} = mg$Given: $I = 25 \,A$,$l = 1 \,m$,$m = 2.5 \,g = 2.5 	imes 10^{-3} \,kg$,$g = 9.8 \,m/s^2$,$\mu_0 = 4 \pi 	imes 10^{-7} \,T \cdot m/A$.Substituting the values:$h = \frac{\mu_0 I^2 l}{2 \pi m g} = \frac{(4 \pi 	imes 10^{-7}) 	imes (25)^2 	imes 1}{2 \pi 	imes (2.5 	imes 10^{-3}) 	imes 9.8}$$h = \frac{2 	imes 10^{-7} 	imes 625}{2.5 	imes 10^{-3} 	imes 9.8} = \frac{1250 	imes 10^{-7}}{24.5 	imes 10^{-3}} = \frac{1250}{24.5} 	imes 10^{-4} \,m \approx 51 	imes 10^{-4} \,m \approx 5.1 \,mm$.Rounding to the nearest given option,the height is $5 \,mm$.

Similar Questions

Currents of $10\, A$ and $2\, A$ are passed through two parallel thin wires $A$ and $B$ respectively in opposite directions. Wire $A$ is infinitely long and the length of wire $B$ is $2\, m$. The force acting on the conductor $B$,which is situated at a distance of $10\, cm$ from $A$,will be:

Two parallel wires $1 \text{ m}$ apart carry currents of $1 \text{ A}$ and $3 \text{ A}$ respectively in opposite directions. The force per unit length acting between these two wires 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:

Three long straight wires $A$,$B$,and $C$ are carrying currents as shown in the figure. Then the resultant force on $B$ is directed .....

$A$ horizontal metallic rod of mass $m$ and length $l$ is supported by two vertical identical springs of spring constant $K$ each and natural length $l_0$. $A$ current $i$ is flowing in the rod in the direction shown. If the rod is in equilibrium,then the length of each spring in this state is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module