A loop carrying current I lies in the x-y plane as shown in the figure- The unit vector hat-k- is coming out of the plane of the paper- The magnetic moment of the current loop isa-2-Ihat-k-displaystyle frac-pi-2-1-a-2-Ihat-k-displaystyle frac-pi-2-1 -a-2-Ihat-k-2pi-1-a-2-Ihat-k-

Question

A loop carrying current I lies in the  x-y plane as shown in the figure- The unit vector hat-k- is coming out of the plane of the paper- The magnetic moment of the current loop isa-2-Ihat-k-displaystyle frac-pi-2-1-a-2-Ihat-k-displaystyle frac-pi-2-1 -a-2-Ihat-k-2pi-1-a-2-Ihat-k-

Toppr Admin · Accepted Answer

Magnetic -xA0-moment -M-I-A-I-xD7-2-xD7-x3C0-4a2-a2-x3C0-2-1-a2I-k

A loop carrying current $I$ lies in the $x$ - $y$ plane as shown in the figure. The unit vector $^k$ is coming out of the plane of the paper. The magnetic moment of the current loop is

$a^{2} I^k$
$(\frac{π}{2} + 1) a^{2} I^k$
$- (\frac{π}{2} + 1) a^{2} I^k$
$(2 π + 1) a^{2} I^k$

Magnetic moment $(M) = I . A = I \times (2 \times \frac{π}{4} a^{2} + a^{2}) = (\frac{π}{2} + 1) a^{2} I^k$

A loop carrying current I lies in the x-y plane as shown in the figure. The unit vector ^k is coming out of the plane of the paper. The magnetic moment of the current loop isa2I^k(π2+1)a2I^k−(π2+1)a2I^k(2π+1)a2I^k

Magnetic moment (M)=I.A=I×(2×π4a2+a2)=(π2+1)a2I^k

A loop carrying current $I$ lies in the $x$ - $y$ plane as shown in the figure. The unit vector $^k$ is coming out of the plane of the paper. The magnetic moment of the current loop is

$a^{2} I^k$
$(\frac{π}{2} + 1) a^{2} I^k$
$- (\frac{π}{2} + 1) a^{2} I^k$
$(2 π + 1) a^{2} I^k$

Magnetic moment $(M) = I . A = I \times (2 \times \frac{π}{4} a^{2} + a^{2}) = (\frac{π}{2} + 1) a^{2} I^k$