Let the rectangular loop be placed in such a way that its plane makes an angle \( \theta \) with the direction of the magnetic field \( \vec{B} \).
Magnetic force on a current-carrying conductor of length \( \vec{l} \) is given by:
\( \vec{F} = I (\vec{l} \times \vec{B}) \)
In a rectangular loop, the opposite sides experience equal and opposite forces, but these forces do not act along the same line. Hence, they form a couple which produces a torque.
Magnitude of Torque:
Let the area of the rectangular loop be:
\( A = l \times b \)
Torque \( \tau \) is given by:
\( \tau = IAB \sin \theta \)
Where:
Define the magnetic moment \( \vec{m} \) of the loop as:
\( \vec{m} = I \vec{A} \)
(Direction of \( \vec{A} \) is given by the right-hand rule perpendicular to the plane of the loop)
Then, torque in vector form is:
\( \vec{\tau} = \vec{m} \times \vec{B} \)
The torque acting on a rectangular current loop placed in a uniform magnetic field is:
\( \tau = IAB \sin \theta \), and in vector form, \( \vec{\tau} = \vec{m} \times \vec{B} \)
In the given reaction sequence, the structure of Y would be: