Current i is flowing in the rectangular loop | vedclass

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

Question

(C) The magnetic moment $\vec \mu$ of a current-carrying loop is given by $\vec \mu = i \vec A$,where $i$ is the current and $\vec A$ is the area vect

Vedclass · Accepted Answer

$iab(\sin 30^\circ \hat i + \cos 30^\circ \hat k)$

Vedclass · Answer

(C) The magnetic moment $\vec \mu$ of a current-carrying loop is given by $\vec \mu = i \vec A$,where $i$ is the current and $\vec A$ is the area vector.The area vector $\vec A$ has a magnitude equal to the area of the loop $(A = a 	imes b)$ and its direction is perpendicular to the plane of the loop.The loop lies in a plane that makes an angle of $30^\circ$ with the $xy$-plane,or more specifically,the normal to the loop makes an angle with the axes.Looking at the geometry,the side of length $b$ is along the $y$-axis. The side of length $a$ makes an angle of $30^\circ$ with the $x$-axis in the $xz$-plane.The normal vector $\vec n$ to the loop can be found by taking the cross product of the two vectors representing the sides of the rectangle.Vector along side $b$ is $\vec b = b \hat j$.Vector along side $a$ is $\vec a = a \cos 30^\circ \hat i + a \sin 30^\circ \hat k$.The area vector $\vec A = \vec a 	imes \vec b = (a \cos 30^\circ \hat i + a \sin 30^\circ \hat k) 	imes (b \hat j)$.Using the cross product rules ($\hat i 	imes \hat j = \hat k$ and $\hat k 	imes \hat j = -\hat i$):$\vec A = ab \cos 30^\circ \hat k - ab \sin 30^\circ \hat i$.Wait,checking the orientation: The magnetic moment direction is determined by the right-hand rule. For the given current direction,the normal vector points in the direction of $(\sin 30^\circ \hat i + \cos 30^\circ \hat k)$.Thus,$\vec \mu = iab(\sin 30^\circ \hat i + \cos 30^\circ \hat k)$.

Current $i$ is flowing in the rectangular loop placed in the $xyz$ plane as shown in the figure. Find the magnetic moment of the loop.

Similar Questions

The magnetic moments associated with two closely wound circular coils $A$ and $B$ of radius $r_A = 10 \ cm$ and $r_B = 20 \ cm$ respectively are equal if: (Where $N_A, I_A$ and $N_B, I_B$ are the number of turns and current of $A$ and $B$ respectively)

$A$ charged particle (charge $q$) is moving in a circle of radius $R$ with uniform speed $v.$ The associated magnetic moment $\mu$ is given by

An electron in a circular orbit of radius $0.05 \ nm$ performs $10^{16}$ revolutions per second. The magnetic moment due to this rotation of the electron is (in $A \ m^{2}$):

The magnetic moment of a current $(I)$ carrying circular coil of radius '$r$' and number of turns '$n$' depends on

In a hydrogen atom,the electron is making $6.6 \times 10^{15} \text{ rev } s^{-1}$ around the nucleus in an orbit of radius $0.528 \text{ Å}$. The magnetic moment in $\text{A-m}^2$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module