A rectangular loop of sides 10 cm and 5 cm | vedclass

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

Question

(B) The magnetic moment $\overrightarrow{M}$ of a current loop is given by $\overrightarrow{M} = I \overrightarrow{A}$,where the direction of area vec

Vedclass · Accepted Answer

$(B)$ and $(D)$,respectively

Vedclass · Answer

(B) The magnetic moment $\overrightarrow{M}$ of a current loop is given by $\overrightarrow{M} = I \overrightarrow{A}$,where the direction of area vector $\overrightarrow{A}$ is determined by the right-hand thumb rule.For stable equilibrium,the magnetic moment $\overrightarrow{M}$ must be parallel to the magnetic field $\overrightarrow{B}$ (i.e.,$	heta = 0^\circ$).For unstable equilibrium,the magnetic moment $\overrightarrow{M}$ must be anti-parallel to the magnetic field $\overrightarrow{B}$ (i.e.,$	heta = 180^\circ$).In figure $(a)$,the current flows in the $yz$-plane. By the right-hand rule,the area vector $\overrightarrow{A}$ points in the $+x$ direction.In figure $(b)$,the current flows in the $xy$-plane. By the right-hand rule,the area vector $\overrightarrow{A}$ points in the $+z$ direction. Since $\overrightarrow{B}$ is in the $+z$ direction,$\overrightarrow{M} \parallel \overrightarrow{B}$,which corresponds to stable equilibrium.In figure $(c)$,the current flows in the $xz$-plane. By the right-hand rule,the area vector $\overrightarrow{A}$ points in the $-y$ direction.In figure $(d)$,the current flows in the $xy$-plane. By the right-hand rule,the area vector $\overrightarrow{A}$ points in the $-z$ direction. Since $\overrightarrow{B}$ is in the $+z$ direction,$\overrightarrow{M}$ is anti-parallel to $\overrightarrow{B}$,which corresponds to unstable equilibrium.Therefore,$(b)$ is stable and $(d)$ is unstable.

Similar Questions

$A$ current-carrying loop is placed in a uniform magnetic field $B$ in different orientations $I$, $II$, $III$, and $IV$ as shown in the figure. The correct order of decreasing potential energy is ($\hat{n}$ is the unit vector normal to the plane of the loop).

$A$ rectangular loop carrying a current $i$ is placed in a uniform magnetic field $B$. The area enclosed by the loop is $A$. If there are $n$ turns in the loop,the torque acting on the loop is given by

$A$ current $i$ flows in a circular coil of radius $r$. If the coil is placed in a uniform magnetic field $B$ with its plane parallel to the field,the magnitude of the torque that acts on the coil is

$A$ single circular loop of radius $1.00 \, m$ carries a current of $10.0 \, mA$. It is placed in a uniform magnetic field of magnitude $0.500 \, T$ that is directed parallel to the plane of the loop as shown in the figure. The magnitude of the torque exerted on the loop by the magnetic field is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module