A rectangular loop carrying a current i is si | vedclass

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

Question

(C) The magnetic field produced by the long straight wire at a distance $r$ is given by $B = \frac{\mu_0 i}{2\pi r}$.For the side of the loop closer t

Vedclass · Accepted Answer

Move towards the wire

Vedclass · Answer

(C) The magnetic field produced by the long straight wire at a distance $r$ is given by $B = \frac{\mu_0 i}{2\pi r}$.For the side of the loop closer to the wire at distance $r_1$,the current flows in the same direction as the wire. According to Fleming's left-hand rule,the force $F_1$ on this side is attractive (towards the wire).For the side of the loop farther from the wire at distance $r_2$,the current flows in the opposite direction to the wire. The force $F_2$ on this side is repulsive (away from the wire).Since $r_1  B_2)$.Consequently,the attractive force $F_1$ is greater than the repulsive force $F_2$ $(F_1 > F_2)$.Therefore,the net force $F_{net} = F_1 - F_2$ acts towards the wire,and the loop will move towards the wire.

$A$ rectangular loop carrying a current $i$ is situated near a long straight wire such that the wire is parallel to one of the sides of the loop and is in the plane of the loop. If a steady current $i$ is established in the wire as shown in the figure,the loop will

Similar Questions

$A$ conducting wire $PQ$ carries a current $10 \ A$ as shown in the figure. It is placed in a uniform magnetic field $5 \ T$ which is acting normally outwards from the paper. The net force experienced by the wire is:

$A$ wire carrying a current $I$ along the positive $x$-axis has length $L$. It is kept in a magnetic field $\overrightarrow{B} = (2\hat{i} + 3\hat{j} - 4\hat{k}) \text{ T}$. The magnitude of the magnetic force acting on the wire is $..........IL$.

$A$ current of $1 \ A$ is flowing along the positive $x$-axis through a straight wire of length $0.5 \ m$ placed in a region of a magnetic field given by $\vec{B} = (2\hat{i} + 4\hat{j}) \ T$. The magnitude and the direction of the force experienced by the wire are,respectively:

The force on a current-carrying loop (radius $= R$) in a uniform magnetic field $(B)$,which is at an angle $30^{\circ}$ with the normal,will be:

$A$ conductor $ABCDE$,shaped as shown,carries a current $i$. It is placed in the $xy$ plane with the ends $A$ and $E$ on the $x$-axis. $A$ uniform magnetic field of magnitude $B$ exists in the region. The force acting on it will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module