Exercise - Magnetic force on an electric charge basic
if (int_count_times_randomized == 0){
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} else {
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return 30;
} else {
return random_min_max_precision(0, 350, -1); //defined in setup_exercise_all.js
}
return (true);
Find the force.
Given:
$q = ~q~C$
$|\vec v| = ~v~ m/s$
$|\vec B| = [[return Math.abs(~b~)]] T$
Find the magnitude of the magnetic force $|\vec F_B|$.
Find the direction of the magnetic force $\theta_F$.
When entering the angle $\theta_F$, use the angle measured from the positive $x$-axis. If the force is into or out of the page, then enter "in" or "out".
namespace_magnetism_basic.phi = ~theta_b~ - ~theta_v~; //In degrees, in [-360, +360].
if (namespace_magnetism_basic.phi < -180){ //>
namespace_magnetism_basic.phi += 360;
} else if (namespace_magnetism_basic.phi > +180){ //>
namespace_magnetism_basic.phi -= 360;
} //Now in [-180, +180].
namespace_magnetism_basic.force = ~q~ * ~v~ * ~b~ * Math.sin(namespace_magnetism_basic.phi * Math.PI/180); //Positive for out of page, negative for into page.
Hint:
$\vec F_B = q \vec v \times \vec B$
$|\vec F_B| = |q v B \sin \phi|$
Solution
From the figure, the angle between $\vec v$ and $\vec B$ is $\phi = [[return Math.abs(namespace_magnetism_basic.phi)]]^\circ$.
$$
\begin{eqnarray}
|\vec F_B| &=& |q v B \sin \phi|
= |(~q~ C)([[return Math.abs(~v~)]] m/s)([[return Math.abs(~b~)]] T) \sin ([[return Math.abs(namespace_magnetism_basic.phi)]]^\circ) | \\
&=& [[return Math.abs(namespace_magnetism_basic.force).toFixed(2)]] N
\end{eqnarray}
$$
Direction of the force is found using the right hand rule.
Find the magnitude of the magnetic force $|\vec F_B|$.
Find the direction of the magnetic force $\theta_F$.
When entering the angle $\theta_F$, use the angle measured from the positive $x$-axis. If the force is into or out of the page, then enter "in" or "out".
From the figure, the angle between $\vec v$ and $\vec B$ is $\phi = 90^\circ$.
$$
\begin{eqnarray}
|\vec F_B| &=& |q v B \sin \phi|
= |(~q~ C)([[return Math.abs(~v~)]] m/s)([[return Math.abs(~b~)]] T) \sin (90^\circ) | \\
&=& [[return Math.abs(namespace_magnetism_basic.force).toFixed(2)]] N
\end{eqnarray}
$$
Direction of the force is found using the right hand rule.
From the figure, the angle between $\vec v$ and $\vec B$ is $\phi = 90^\circ$.
$$
\begin{eqnarray}
|\vec F_B| &=& |q v B \sin \phi| = |q v B \sin 90^\circ| = |q|vB \\
\Rightarrow B &=& \frac{|\vec F_B|}{|q|v}
= \frac{[[return Math.abs(namespace_magnetism_basic.force).toFixed(2)]] N}{|~q~ C|([[return Math.abs(~v~)]] m/s)}
= [[return Math.abs(~b~)]] T
\end{eqnarray}
$$
Direction of the field is found using the right hand rule.
$|\vec B| = $
return Math.abs(~b~)
5%
$T$ Direction of magnetic field:
return (~b~>0)? "out": "in";
not_number
(enter "in" or "out")
magnetism || force_magnetic
Exercise - Circular trajectory in a uniform magnetic field
Find the magnitude of the magnetic force $|\vec F_B|$.
Find the radius of the trajectory.
Does the charge move clockwise (CW) or counter-clockwise (CCW)?
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return -2;
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while (Math.abs(q) < 2){ //>
q = random_min_max_precision(-3, 3, 1);
}
return q;
}
if (int_count_times_randomized == 0){
return 3;
} else {
return random_min_max_precision(2, 4, 1); //defined in setup_exercise_all.js
}
return 90;
if (int_count_times_randomized == 0){
return 1.5;
} else {
let b = random_min_max_precision(-2, 2, 1, true);
while (Math.abs(b) < 1.5){ //>
b = random_min_max_precision(-2, 2, 1, true);
}
return b;
}
if (int_count_times_randomized == 0){
return 2.5;
} else {
return random_min_max_precision(1, 3, 1); //defined in setup_exercise_all.js
}
Hint:
$F_{centripetal} = \frac{mv^2}{r}$
$\vec F_B = q \vec v \times \vec B$
$|\vec F_B| = |q v B \sin \phi|$
$F_{centripetal} = F_B$ for circular trajectory
Solution
Final solution.
From the figure, the angle between $\vec v$ and $\vec B$ is $\phi = 90^\circ$.
$$
\begin{eqnarray}
|\vec F_B| &=& |q v B \sin \phi| = |q v B \sin 90^\circ| = |q| vB \\
&=& |~q~ C|([[return Math.abs(~v~)]] m/s)([[return Math.abs(~b~)]] T) \\
&=& [[return Math.abs(namespace_magnetism_basic.force).toFixed(2)]] N
\end{eqnarray}
$$
The magnetic force above must equal to the centripetal force $F_{centripetal} = \frac{m v^2}{r}$:
$$
\begin{eqnarray}
\frac{m v^2}{r} &=& |q| v B \\
\Rightarrow r &=& \frac{mv}{|q|B}
= \frac{(~m~ kg)([[return Math.abs(~v~)]] m/s)}{|~q~ C| ([[return Math.abs(~b~)]] T)} \\
&=& [[return Math.abs(namespace_magnetism_basic.r).toFixed(2)]] m
\end{eqnarray}
$$
Direction of the force is found using the right hand rule.
An electric potential $V$ accelerates the charge from rest.
A charge is accelerated from rest by an electric potential difference $\Delta V$ in the lower region before entering the upper region with a uniform magnetic field.
Given:
$|\Delta V| = V = [[return Math.abs(namespace_magnetism_basic.delta_V).toFixed(2)]] V$
$q = ~q~C$
$|\vec B| = [[return Math.abs(~b~)]] T$
$m = ~m~ kg$
Find the work $W_E$ done by the electric field on the charge in the lower region.
Find the speed $v$ of the charge when it enters the region of the magnetic field.
Find the magnitude of the magnetic force $|\vec F_B|$.
Find the radius of the trajectory.
Does the charge move clockwise (CW) or counter-clockwise (CCW) in the upper region?
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return -2;
} else {
let q = random_min_max_precision(-3, 3, 1);
while (Math.abs(q) < 2){ //>
q = random_min_max_precision(-3, 3, 1);
}
return q;
}
if (int_count_times_randomized == 0){
return 3;
} else {
return random_min_max_precision(2, 4, 1); //defined in setup_exercise_all.js
}
return 90;
if (int_count_times_randomized == 0){
return 1.5;
} else {
let b = random_min_max_precision(-2, 2, 1, true);
while (Math.abs(b) < 1.5){ //>
b = random_min_max_precision(-2, 2, 1, true);
}
return b;
}
if (int_count_times_randomized == 0){
return 2.5;
} else {
return random_min_max_precision(1, 3, 1); //defined in setup_exercise_all.js
}
Hint:
$W_E = -q \Delta V = \frac{1}{2}mv^2$
$F_{centripetal} = \frac{mv^2}{r}$
$\vec F_B = q \vec v \times \vec B$
$|\vec F_B| = |q v B \sin \phi|$
$F_{centripetal} = F_B$ for circular trajectory
Solution
[[
if (namespace_magnetism_basic.delta_V>0){
return `
The change in potential $\\Delta V$ as the charges moves through the lower region is $\\Delta V = V = ${(namespace_magnetism_basic.delta_V).toFixed(2)} V$.
Note that $\\Delta V$ is positive because the charge travels from the negative terminal (low $V$) to the postive terminal (high $V$), therefore experence an increase in potential.
`;
} else {
return `
The change in potential $\\Delta V$ as the charges moves through the lower region is $\\Delta V = - V = ${(namespace_magnetism_basic.delta_V).toFixed(2)} V$.
Note that $\\Delta V$ is negative because the charge travels from the positive terminal (high $V$) to the negative terminal (low $V$), therefore experence a decrease in potential.
`;
}
]]
The work done on a charge moving through an electric potential difference $\Delta V$ is:
$$
W_E = -q \Delta V
= -(~q~ C) ([[return (namespace_magnetism_basic.delta_V).toFixed(2)]] V)
= [[return (namespace_magnetism_basic.ke).toFixed(2)]] J
$$
Alternatively, we could have just taken the absolute value on both sides and write:
$$
W_E = |-q \Delta V | = |q| V
= ([[return Math.abs(~q~)]] C) ([[return Math.abs(namespace_magnetism_basic.delta_V).toFixed(2)]] V)
= [[return (namespace_magnetism_basic.ke).toFixed(2)]] J
$$
This gives the same result. We could do this because we know the potential is accelerating the charge, hence giving it energy. This means $W_E$ must be positive, so all the minus signs on the right must cancel out in the end. If so there is no point keeping track of all the negative signs.
The work is the energy transferred to the charge by the E-field, becoming the charge's kinetic energy:
$$
\begin{eqnarray}
\frac{1}{2}mv^2 &=& W_E \\
\Rightarrow v &=& \sqrt{\frac{2W_E}{m}} \\
&=& \sqrt{\frac{2([[return (namespace_magnetism_basic.ke).toFixed(2)]] J)}{~m~ kg}}
= ~v~ m/s
\end{eqnarray}
$$
Final solution.
In the upper region, the angle between $\vec v$ and $\vec B$ is $\phi = 90^\circ$.
$$
\begin{eqnarray}
|\vec F_B| &=& |q v B \sin \phi| = |q v B \sin 90^\circ| = |q| vB \\
&=& |~q~ C|([[return Math.abs(~v~)]] m/s)([[return Math.abs(~b~)]] T) \\
&=& [[return Math.abs(namespace_magnetism_basic.force).toFixed(2)]] N
\end{eqnarray}
$$
The magnetic force above must equal to the centripetal force $F_{centripetal} = \frac{m v^2}{r}$:
$$
\begin{eqnarray}
\frac{m v^2}{r} &=& |q| v B \\
\Rightarrow r &=& \frac{mv}{|q|B}
= \frac{(~m~ kg)([[return Math.abs(~v~)]] m/s)}{|~q~ C| ([[return Math.abs(~b~)]] T)} \\
&=& [[return Math.abs(namespace_magnetism_basic.r).toFixed(2)]] m
\end{eqnarray}
$$
Direction of the force is found using the right hand rule.
An electric potential $V$ accelerates the charge from rest.
A charge is accelerated from rest by an electric potential difference $\Delta V$ in the lower region before entering the upper region with a uniform magnetic field.
Given:
$|\Delta V| = V = [[return Math.abs(namespace_magnetism_basic.delta_V).toFixed(2)]] V$
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let q = random_min_max_precision(-3, 3, 1);
while (Math.abs(q) < 2){ //>
q = random_min_max_precision(-3, 3, 1);
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return q;
}
if (int_count_times_randomized == 0){
return 3.5;
} else {
return random_min_max_precision(2, 4, 1); //defined in setup_exercise_all.js
}
return 90;
if (int_count_times_randomized == 0){
return 1.5;
} else {
let b = random_min_max_precision(-2, 2, 1, true);
while (Math.abs(b) < 1.5){ //>
b = random_min_max_precision(-2, 2, 1, true);
}
return b;
}
if (int_count_times_randomized == 0){
return 2.5;
} else {
return random_min_max_precision(1, 3, 1); //defined in setup_exercise_all.js
}
Hint:
$W_E = -q \Delta V = \frac{1}{2}mv^2$
$F_{centripetal} = \frac{mv^2}{r}$
$\vec F_B = q \vec v \times \vec B$
$|\vec F_B| = |q v B \sin \phi|$
$F_{centripetal} = F_B$ for circular trajectory
Solution
Work done by the E-field is $W_E = -q \Delta V > 0$. Taking the absolute value on both sides:
$$
W_E = |-q \Delta V | = |q| V
$$
We could do this because we know the potential is accelerating the charge, hence giving it energy. This means $W_E$ must be positive, so all the minus signs on the right must cancel out in the end. If so there is no point keeping track of all the negative signs.
The work is the energy transferred to the charge by the E-field, becoming the charge's kinetic energy:
$$
\begin{eqnarray}
\frac{1}{2}mv^2 &=& W_E = |q|V \\
\Rightarrow v &=& \sqrt{\frac{2|q| V}{m}}
\end{eqnarray}
$$
Final solution.
In the upper region, the angle between $\vec v$ and $\vec B$ is $\phi = 90^\circ$.
$$
\begin{eqnarray}
|\vec F_B| &=& |q v B \sin \phi| = |q v B \sin 90^\circ| = |q| vB
\end{eqnarray}
$$
The magnetic force above must equal to the centripetal force $F_{centripetal} = \frac{m v^2}{r}$:
$$
\begin{eqnarray}
\frac{m v^2}{r} &=& |q| v B \\
\Rightarrow v &=& \frac{|q|B r}{m}
\end{eqnarray}
$$
Squaring both sides and putting in $v = \sqrt{\frac{2|q| V}{m}}$ that we found earlier:
$$
\begin{eqnarray}
\frac{2|q| V}{m} &=& \frac{|q|^2 B^2 r^2}{m^2} \\
\Rightarrow \frac{|q|}{m} &=& \frac{2 V }{B^2 r^2}
= \frac{2 ([[return Math.abs(namespace_magnetism_basic.delta_V).toFixed(2)]] V) }{([[return Math.abs(~b~)]] T)^2 ([[return Math.abs(namespace_magnetism_basic.r).toFixed(2)]] m)^2} \\
&=& [[return Math.abs(namespace_magnetism_basic.q_to_m).toFixed(2)]] C/kg
\end{eqnarray}
$$
We found $\frac{|q|}{m}$ but not $\frac{q}{m}$, so we still have to determine the sign of $q$ by applying the right hand rule to the figure. This gives the final answer:
$$
\frac{q}{m} = [[return (namespace_magnetism_basic.q_to_m).toFixed(2)]] C/kg
$$
Find the magnitude and direction of the electric force $\vec F_E$.
Find the magnitude and direction of the magnetic force $\vec F_B$.
Find the speed of the charge $v$.
if (int_count_times_randomized == 0){
return -2.5;
} else {
let q = random_min_max_precision(-3, 3, 1);
while (Math.abs(q) < 2){ //>
q = random_min_max_precision(-3, 3, 1);
}
return q;
}
if (int_count_times_randomized == 0){
return 3.5;
} else {
return random_min_max_precision(2, 4, 1); //defined in setup_exercise_all.js
}
return 0;
if (int_count_times_randomized == 0){
return 1.5;
} else {
let b = random_min_max_precision(-2, 2, 1, true);
while (Math.abs(b) < 1.5){ //>
b = random_min_max_precision(-2, 2, 1, true);
}
return b;
}
Hint:
$\vec F_E = q \vec E$
$\vec F_B = q \vec v \times \vec B$
$|\vec F_B| = |q v B \sin \phi|$
$|\vec F_B| = |\vec F_E|$ when undeflected
Solution
Final solution.
The electric force:
$$
\begin{eqnarray}
\vec F_E &=& q \vec E \\
\Rightarrow |\vec F_E| &=& |q| |\vec E|
= ([[return Math.abs(~q~)]] C)([[return (namespace_magnetism_basic.e_field).toFixed(2)]] N/C) \\
&=& [[return Math.abs(namespace_magnetism_basic.force).toFixed(2)]] N
\end{eqnarray}
$$
[[
if (~q~>0){
return `
$\\vec F_E$ points in the same direction as $\\vec E$ when $q \\gt 0$.
`;
} else {
return `
$\\vec F_E$ points in the opposite direction as $\\vec E$ when $q \\lt 0$.
`;
}
]]
Since the charge moves undeflected (as stated in the question), the magnetic force must cancel the electric force. It means $\vec F_B$ has the same magnitude as $\vec F_E$ but points in the opposite direction.
The magnetic force:
$$
\begin{eqnarray}
|\vec F_B| &=& [[return Math.abs(namespace_magnetism_basic.force).toFixed(2)]] N
\end{eqnarray}
$$
[[
if (namespace_magnetism_basic.force>0){
return `
Since $\\vec F_E$ points up, $\\vec F_B$ must point down.
`;
} else {
return `
Since $\\vec F_E$ points down, $\\vec F_B$ must point up.
`;
}
]]
To find the velocity, set $|\vec F_B| = |\vec F_E|$:
$$
\begin{eqnarray}
|q| v B \sin 90^\circ &=& |q| E \\
\Rightarrow v &=& \frac{E}{B}
= \frac{[[return (namespace_magnetism_basic.e_field).toFixed(2)]] N/C }{[[return Math.abs(~b~).toFixed(2)]] T} \\
&=& ~v~ m/s
\end{eqnarray}
$$
A wire of length $L$ with current $I$ passes inside a magnetic field.
Given:
$I = ~i~ A$
$L = ~l~ m$
$|\vec B| = [[return Math.abs(~b~)]] T$
Find the magnitude of the magnetic force $|\vec F_B|$.
Find the direction of the magnetic force $\theta_F$.
When entering the angle $\theta_F$, use the angle measured from the positive $x$-axis. If the force is into or out of the page, then enter "in" or "out".
if (int_count_times_randomized == 0){
return 2.2;
} else {
return random_min_max_precision(1.1, 3, 1); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 0;
} else {
return random_min_max_precision(0, 350, -1); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 1.3;
} else {
return random_min_max_precision(0.2, 2, 1); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 2;
} else {
let b = random_min_max_precision(-2, 2, 1);
while (Math.abs(b)<0.5){ //>
b = random_min_max_precision(-2, 2, 1);
}
return b;
}
Hint:
$\vec F_B = I \vec L \times \vec B$
$|\vec F_B| = |I w B \sin \phi|$
Solution
Final solution.
From the figure, the angle between $\vec v$ and $\vec B$ is $\phi = 90^\circ$.
$$
\begin{eqnarray}
|\vec F_B| &=& |I w B \sin \phi|
= (~i~ A)(~l~ m)([[return Math.abs(~b~)]] T) \sin (90^\circ) \\
&=& [[return Math.abs(namespace_magnetism_basic.force).toFixed(2)]] N
\end{eqnarray}
$$
Direction of the force is found using the right hand rule.
Find the magnitude of the magnetic force $|\vec F_B|$.
Find the direction of the magnetic force $\theta_F$.
When entering the angle $\theta_F$, use the angle measured from the positive $x$-axis. If the force is into or out of the page, then enter "in" or "out".
From the figure, the angle between $\vec v$ and $\vec B$ is $\phi = 90^\circ$.
$$
\begin{eqnarray}
|\vec F_B| &=& |I w B \sin \phi|
= |(~i~ A)(~l~ m)([[return Math.abs(~b~)]] T) \sin (90^\circ)| \\
&=& [[return Math.abs(namespace_magnetism_basic.force).toFixed(2)]] N
\end{eqnarray}
$$
Direction of the force is found using the right hand rule.
$^\circ$ (enter "in", "out", or "no force" if the force points into or out of the page, or if $|\vec F_B| = 0N$)
magnetism || force_magnetic || current
Example - Torque on a loop of current
A loop carrying a current inside a magnetic field. The $x$-axis points to the right (along $\vec B$), the $y$-axis points into the screen, and the $z$-axis points vertically up.
A rectangular loop of wire carrying a current $I$ inside a uniform magnetic field $\vec B$ pointing along the $x$-axis. The area vector $\vec A$ makes an angle $\theta$ with $\vec B$ as shown. The sides of the rectangle are labeled 1, 2, 3, and 4. The width and the height of the rectangle are $w$ and $h$ respectively.
Calculate the force vectors on each side of the loop.
Calculate the torque vectors exerted by each side about the center of mass of the loop.
Calculate the total torque vector on the loop in terms of $\vec \mu = I \vec A$.
Is the torque causing a clockwise or counter-clockwise rotation (viewed from above)?
Solution
The view of the loop from above.
Wire 1
$\vec L_1 = + h \hat k$, so the force on wire 1 is:
$$
\begin{eqnarray}
\vec F_1 &=& I \vec L_1 \times \vec B = I (h \hat k) \times (B \hat i) \\
&=& IhB \hat j
\end{eqnarray}
$$
Wire 2
Note that $\vec L_2$ points southeast, $\vec L_2 = w \sin \theta \hat i - w \cos \theta \hat j$. The force on wire 2 is:
$$
\begin{eqnarray}
\vec F_2 &=& I \vec L_2 \times \vec B = I (w \sin \theta \hat i - w \cos \theta \hat j) \times (B \hat i) \\
&=& I w B (\sin \theta \hat i \times B \hat i - \cos \theta \hat j \times B \hat i) \\
&=& I w B \cos\theta \hat k
\end{eqnarray}
$$
Wire 3
$\vec L_3 = - h \hat k$, so the force on wire 3 is:
$$
\begin{eqnarray}
\vec F_3 &=& I \vec L_3 \times \vec B = I (-h \hat k) \times (B \hat i) \\
&=& - I h B \hat j
\end{eqnarray}
$$
Wire 4
Note that $\vec L_4$ points northwest, $\vec L_4 = -w \sin \theta \hat i + w \cos \theta \hat j$. The force on wire 4 is:
$$
\begin{eqnarray}
\vec F_4 &=& I \vec L_4 \times \vec B = I (-w \sin \theta \hat i + w \cos \theta \hat j) \times (B \hat i) \\
&=& I w B ( - \sin \theta \hat i \times B \hat i + \cos \theta \hat j \times B \hat i) \\
&=& - I w B \cos\theta \hat k
\end{eqnarray}
$$
Torque on wire 1
We will use $\tau = \vec r \times \vec F$ to compute the torques. Since $\vec r_1$ points northwest (from the middle of the loop to wire 1), we have $\vec r_1 = \frac{w}{2}(-\sin \theta \hat i + \cos \theta \hat j)$.
$$
\begin{eqnarray}
\vec \tau_1 &=& \vec r_1 \times \vec F_1 \\
&=& \frac{w}{2}(-\sin \theta \hat i + \cos \theta \hat j) \times (IhB \hat j) \\
&=& -\frac{1}{2} (I hw B \sin \theta \hat k) \\
&=& -\frac{1}{2} (I A B \sin \theta \hat k) \qquad \text{where $A = hw$} \\
&=& \frac{1}{2} I \vec A \times \vec B
\end{eqnarray}
$$
Torque on wire 3
Similarly $\vec r_3$ points southeast (from the middle of the loop to wire 3), we have $\vec r_3 = \frac{w}{2}(\sin \theta \hat i - \cos \theta \hat j)$.
$$
\begin{eqnarray}
\vec \tau_3 &=& \vec r_3 \times \vec F_3 \\
&=& \frac{w}{2}(\sin \theta \hat i - \cos \theta \hat j) \times (-IhB \hat j) \\
&=& -\frac{1}{2} (I hw B \sin \theta \hat k) \\
&=& -\frac{1}{2} (I A B \sin \theta \hat k) \\
&=& \frac{1}{2} I \vec A \times \vec B
\end{eqnarray}
$$
Torque on wire 2 and 4
The torque for the top and bottom wire is zero because $\vec r_2 = +\frac{h}{2} \hat k$ and $\vec r_4 = -\frac{h}{2} \hat k$ so $\vec r$ and $\vec F$ are parallel / anti-parallel, which gives $\tau = \vec r \times \vec F = \vec 0$.
Total torque
$$
\begin{eqnarray}
\vec \tau &=& \vec \tau_1 + \vec \tau_2 + \vec \tau_3 + \vec \tau_4 \\
&=& \frac{1}{2} I \vec A \times \vec B + \vec 0 + \frac{1}{2} I \vec A \times \vec B + \vec 0 \\
&=& I \vec A \times \vec B \\
&=& \vec \mu \times \vec B
\end{eqnarray}
$$
$\vec \mu = I \vec A$ is known as the magnetic dipole moment.
Rotation
In the moment as shown in the figure, the torque is turning the loop clockwise. In general $\tau = \vec \mu \times \vec B$ causes the magnetic dipole $\vec \mu$ to rotate itself to align with $\vec B$.
Application
By cleverly switching the direction of the current, the torque you just computed can be arranged to always turn the loop in one direction. This is the basis of many motors.
A motor works by changing the direction of the current so the torque always rotates the loop in one direction.
Lookang many thanks to Fu-Kwun Hwang and author of Easy Java Simulation = Francisco Esquembre, CC BY-SA 3.0 , via Wikimedia Commons
magnetism || force_magnetic
Example - Drawing magnetic force on an electric charge
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