- #1

sergiokapone

- 302

- 17

## Homework Statement

The wire loop ##ABCD## is in a magnetic field of the long thin rod magnet with a magnetic moment per unit volume of ##\mathfrak{M}_0## and a cross section ##S##. The north pole of this magnet is in the center of the loop, and the magnet is perpendicular to the plane of the loop. Emf is connected to the opposite edges of the diagonal ##AC## , thus the current I is flow along sides of the circuit. Find the torque of force couple M and its direction.

## Homework Equations

[/B]

## The Attempt at a Solution

**The magnet field of the long magnet near the north pole one can represent as magnetic field of monopole**

\begin{equation}

H = \frac{q_m}{r^2}

\end{equation}

\begin{equation}

H = \frac{q_m}{r^2}

\end{equation}

where ##q_m## -- effective magnetic charge, which one can find from definition of magnetic momentum ##p_m = q_ml##:

\begin{equation*}

q_ml = \mathfrak{M}_0 S l

\end{equation*}

thus, the effective magnetic charge is

\begin{equation}

q_m = \mathfrak{M}_0 S.

\end{equation}

The Ampere Force acting on sides of wire loop:

\begin{equation}

dF = Idl H \sin\alpha,

\end{equation}

And, so the torque is

\begin{equation}

M = \int rdF = I\mathfrak{M}_0 S \int \frac{\sin\alpha}{r} dl

\end{equation}

from the geometry ##dl\sin\alpha = r d\alpha##, thus

where ##q_m## -- effective magnetic charge, which one can find from definition of magnetic momentum ##p_m = q_ml##:

\begin{equation*}

q_ml = \mathfrak{M}_0 S l

\end{equation*}

thus, the effective magnetic charge is

\begin{equation}

q_m = \mathfrak{M}_0 S.

\end{equation}

The Ampere Force acting on sides of wire loop:

\begin{equation}

dF = Idl H \sin\alpha,

\end{equation}

And, so the torque is

\begin{equation}

M = \int rdF = I\mathfrak{M}_0 S \int \frac{\sin\alpha}{r} dl

\end{equation}

from the geometry ##dl\sin\alpha = r d\alpha##, thus

**\begin{equation}**

M = \int rdF = I\mathfrak{M}_0 S 4 \frac{\pi}{2}

\end{equation}

The answer in the problems book is

\begin{equation}

M = \int rdF = 4 I\mathfrak{M}_0 S

\end{equation}

Where did I go wrong?

M = \int rdF = I\mathfrak{M}_0 S 4 \frac{\pi}{2}

\end{equation}

The answer in the problems book is

\begin{equation}

M = \int rdF = 4 I\mathfrak{M}_0 S

\end{equation}

Where did I go wrong?