Q165 / 180 · COMEDK 2024 (Morning Shift) · Physics
A circular coil of radius $0.1 \mathrm{~m}$ is placed in the $\mathrm{X}-\mathrm{Y}$ plane and a current $2 \mathrm{~A}$ is passed through the coil in the clockwise direction when looking from above. Find the magnetic dipole moment of the current loop
- A$0.02 \pi \mathrm{Am}^2 \text { in the }+ \text { ve } \mathrm{Z}-\text { direction }$
- B$0.02 \pi \mathrm{Am}^2 \text { in the }+ \text { ve } \mathrm{Y}-\text { direction }$
- C$0.02 \pi \mathrm{Am}^2 \text { in the -ve } \mathrm{X} \text { - direction }$
- Correct answerD$0.02 \pi \mathrm{Am}^2 \text { in the }-\mathrm{ve} \mathrm{~Z}-\text { direction }$
Solution
The magnetic dipole moment $\vec{m}$ of a current loop is given by the formula $\vec{m} = I \vec{A}$, where $I$ is the current and $\vec{A}$ is the area vector.
The radius of the coil is $r = 0.1 \text{ m}$. The area of the circular coil is $A = \pi r^2 = \pi (0.1)^2 = 0.01 \pi \text{ m}^2$.
The current $I = 2 \text{ A}$. The magnitude of the magnetic dipole moment is $m = I A = 2 \times 0.01 \pi = 0.02 \pi \text{ Am}^2$.
The coil is placed in the X-Y plane. According to the right-hand rule, for a clockwise current when viewed from above (looking down the positive Z-axis), the direction of the area vector $\vec{A}$ is along the negative Z-axis. Therefore, the magnetic dipole moment $\vec{m}$ is directed in the negative Z-direction.
Answer: $0.02 \pi \mathrm{Am}^2 \text { in the }-\mathrm{ve} \mathrm{~Z}-\text { direction }$
The radius of the coil is $r = 0.1 \text{ m}$. The area of the circular coil is $A = \pi r^2 = \pi (0.1)^2 = 0.01 \pi \text{ m}^2$.
The current $I = 2 \text{ A}$. The magnitude of the magnetic dipole moment is $m = I A = 2 \times 0.01 \pi = 0.02 \pi \text{ Am}^2$.
The coil is placed in the X-Y plane. According to the right-hand rule, for a clockwise current when viewed from above (looking down the positive Z-axis), the direction of the area vector $\vec{A}$ is along the negative Z-axis. Therefore, the magnetic dipole moment $\vec{m}$ is directed in the negative Z-direction.
Answer: $0.02 \pi \mathrm{Am}^2 \text { in the }-\mathrm{ve} \mathrm{~Z}-\text { direction }$
Hence, the correct answer is option: 'D'.