Solution

We first calculate the area in $m^2$ using $r = ~r~ cm = [[return (~r~*1e-2)]] m$: $$ A = \pi r^2 = \pi ([[return (~r~*1e-2)]] m)^2 = [[return namespace_em_induction.get_scientific_notation_latex(namespace_em_induction.area, 2)]] m^2 $$ The angle between $\vec A$ and $\vec B_{initial}$ is $[[return namespace_em_induction.theta_i]] ^\circ$. The initial the magnetic flux is: $$ \begin{eqnarray} \Phi_{initial} &=& N B A \cos \theta_{initial} = (~N~) ([[return Math.abs(~b_i~)]] T)([[return namespace_em_induction.get_scientific_notation_latex(namespace_em_induction.area, 2)]] m^2) (\cos([[return namespace_em_induction.theta_i]] ^\circ)) \\ &=& [[return namespace_em_induction.get_scientific_notation_latex(namespace_em_induction.flux_i, 2)]] Wb \end{eqnarray} $$ The angle between $\vec A$ and $\vec B_{final}$ is $[[return namespace_em_induction.theta_f]] ^\circ$. The final the magnetic flux is: $$ \begin{eqnarray} \Phi_{final} &=& N B A \cos \theta_{final} = (~N~) ([[return Math.abs(~b_f~)]] T)([[return namespace_em_induction.get_scientific_notation_latex(namespace_em_induction.area, 2)]] m^2) (\cos([[return namespace_em_induction.theta_f]] ^\circ)) \\ &=& [[return namespace_em_induction.get_scientific_notation_latex(namespace_em_induction.flux_f, 2)]] Wb \end{eqnarray} $$ The change in the magnetic flux is: $$ \begin{eqnarray} \Delta \Phi &=& \Phi_{final} - \Phi_{initial} = [[return namespace_em_induction.get_scientific_notation_latex(namespace_em_induction.flux_f, 2)]] Wb - ([[return namespace_em_induction.get_scientific_notation_latex(namespace_em_induction.flux_i, 2)]]Wb) \\ &=& [[return namespace_em_induction.get_scientific_notation_latex(namespace_em_induction.delta_flux, 2)]] Wb \end{eqnarray} $$ Applying Faraday's Law: $$ \begin{eqnarray} \mathcal{E} &=& -\frac{d\Phi}{dt} = -\frac{\Delta \Phi}{\Delta t} \\ &=& -\frac{[[return namespace_em_induction.get_scientific_notation_latex(namespace_em_induction.delta_flux, 2)]] Wb}{~t~ s} \\ &=& [[return namespace_em_induction.get_scientific_notation_latex(namespace_em_induction.emf, 2)]] V \end{eqnarray} $$ The direction of the emf can be found using the right hand grip rule. Point your thumb along $\vec A$, then your four fingers wraps along the positive direction for $\mathcal{E}$.

• If $\mathcal{E} \gt 0$: emf is along the four fingers, counterclockwise in this problem.
• If $\mathcal{E} \lt 0$: emf is opposite to the four fingers, clockwise in this problem.

Solution

Take the $z$-axis to point out of the screen as stated in the question, we have:

• The $z$-component of the initial magnetic field is $B_{i, z} = ~b_i~ T$.
• The $z$-component of the final magnetic field is $B_{f, z} = (~b_i~ T) + (~b_delta~ T) = [[return sf_latex(namespace_em_induction.b_f)]] T$.

The area of the ring

$$ \begin{eqnarray} A &=& \pi r^2 \\ &=& \pi (~r~ \times 0.01 m)^2 \\ &=& [[return sf_latex(namespace_em_induction.area)]] m^2 \end{eqnarray} $$

Initial flux

$$ \begin{eqnarray} \Phi_i &=& \vec B_i \cdot \vec A \\ &=& B_{i, z} A & \text{ because $\vec A$ is along the $z$ direction.} \\ &=& (~b_i~ T) ([[return sf_latex(namespace_em_induction.area)]] m^2) \\ &=& [[return sf_latex(namespace_em_induction.flux_i)]] Wb \end{eqnarray} $$

Final flux

$$ \begin{eqnarray} \Phi_f &=& \vec B_f \cdot \vec A \\ &=& B_{f, z} A \\ &=& ([[return sf_latex(namespace_em_induction.b_f)]] T) ([[return sf_latex(namespace_em_induction.area)]] m^2) \\ &=& [[return sf_latex(namespace_em_induction.flux_f)]] Wb \end{eqnarray} $$

Change in flux

$$ \begin{eqnarray} \Delta \Phi_B &=& \Phi_f - \Phi_i \\ &=& ([[return sf_latex(namespace_em_induction.flux_f)]] Wb) - ([[return sf_latex(namespace_em_induction.flux_i)]] Wb) \\ &=& [[return sf_latex(namespace_em_induction.flux_delta)]] Wb \end{eqnarray} $$

Emf

$$ \begin{eqnarray} \mathcal{E} &=& -\frac{d\Phi_B}{dt} \\ &=& -\frac{\Delta \Phi_B}{\Delta t} \\ &=& -\frac{[[return sf_latex(namespace_em_induction.flux_delta)]] Wb}{~t~ s} \\ &=& [[return sf_latex(namespace_em_induction.emf)]] V \end{eqnarray} $$

Magnitude of E field

$$ \begin{eqnarray} \mathcal{E} &=& \oint \vec E \cdot d\vec s \\ \Rightarrow E (2\pi r) &=& |\mathcal{E}| \\ \Rightarrow E &=& \frac{|\mathcal{E}|}{2\pi r} \\ &=& \frac{[[return sf_latex(Math.abs(namespace_em_induction.emf))]] V}{2\pi (~r~ \times 0.01 m)} \\ &=& \frac{[[return sf_latex( Math.abs(namespace_em_induction.emf) )]] V}{[[return sf_latex(namespace_em_induction.circumference)]]m} \\ &=& [[return sf_latex( namespace_em_induction.e_field_abs)]] V/m \end{eqnarray} $$

Direction

The right hand grib rule when applied to an area vector pointing out the screen means:

• $\mathcal{E}\gt 0$: counterclockwise.
• $\mathcal{E}\lt 0$: clockwise.

According to our calculations above, $\mathcal{E}$ is [[return (namespace_em_induction.emf>0)?"greater":"less"]] than zero, it means the emf and the E field must point in the [[return (namespace_em_induction.emf>0)?"counterclockwise":"clockwise"]] direction.

Solution

First we find the area of the capacitor plate: $$ \begin{eqnarray} A_{plate} &=& \pi R^2 \\ &=& \pi (~r_capacitor~ \times 0.01 m )^2 \\ &=& [[return sf_latex(namespace_em_induction.area_capacitor)]] m^2 \end{eqnarray} $$ For comparison, the area of the Amperian loop is: $$ \begin{eqnarray} A_{loop} &=& \pi r^2 \\ &=& \pi (~r_loop~ \times 0.01 m )^2 \\ &=& [[return sf_latex(namespace_em_induction.area_loop)]] m^2 \end{eqnarray} $$

The charge density on the plate is: $$ \begin{eqnarray} \sigma &=& \frac{q}{A_{plate}} \\ &=& \frac{~q~ \times 10^{-9} C}{[[return sf_latex(namespace_em_induction.area_capacitor)]] m^2} \\ &=& [[return sf_latex(namespace_em_induction.charge_density)]] C/m^2 \end{eqnarray} $$

The E field: $$ \begin{eqnarray} E &=& \frac{\sigma}{\epsilon_0} \\ &=& \frac{[[return sf_latex(namespace_em_induction.charge_density)]] C/m^2}{8.854 \times 10^{-12}F/m} \\ &=& [[return sf_latex(namespace_em_induction.e_field)]] V/m \end{eqnarray} $$

$I_C=0A$ because no charges are flowing pass the loop, so $I_{enclosed} = I_D = [[return sf_latex(namespace_em_induction.current_displacement)]] A$.

Applying Ampere-Maxwell law gives: $$ \begin{eqnarray} 2\pi r B &=& \mu_0I_{enclosed} \\ \Rightarrow B &=& \frac{\mu_0 I_{enclosed}}{2\pi r} \\ &=& \frac{(4\pi \times 10^{-7} H/m) ([[return sf_latex(namespace_em_induction.current_displacement)]] A)}{2\pi (~r_loop~ \times 0.01m)} \\ &=& [[return sf_latex(namespace_em_induction.b_field)]] T \end{eqnarray} $$

For comparison, the magnetic field around the wire left of the plate (where $I_C = I$, $I_D = 0$) is: $$ \begin{eqnarray} B_{wire} &=& \frac{\mu_0 I}{2\pi r} \\ &=& \frac{(4\pi \times 10^{-7} H/m) (~i~ A)}{2\pi (~r_loop~ \times 0.01m)} \\ &=& [[return sf_latex(namespace_em_induction.b_field_wire)]] T \end{eqnarray} $$

Solution

For the figure we can see the field is pointing [[return (~b~ > 0)? "out of":"into"]] the screen. Defining the $z$-direction to be pointing out the screen, we have $B_z = ~b~ T$.

Faraday's law of electromagnetic induction: $$ \begin{eqnarray} \mathcal{E} &=& - \frac{d\Phi_B}{dt} \\ &=& - \frac{d(B_z A)}{dt} \\ &=& - B_z \frac{dA}{dt} & \text{magnetic field is constant but area changes} \end{eqnarray} $$

The area inside the loop is $A = l x$, therefore: $$ \begin{eqnarray} \frac{dA}{dt} &=& \frac{d (lx)}{dt} \\ &=& l\frac{dx}{dt} \\ &=& l v \\ &=& (~l~ m) ( ~v~ m/s) \\ &=& [[return sf_latex(namespace_em_induction.da_by_dt)]] m^2/s \end{eqnarray} $$

Putting together: $$ \begin{eqnarray} \mathcal{E} &=& - B_z \frac{dA}{dt} \\ &=& - (~b~ T) ([[return sf_latex(namespace_em_induction.da_by_dt)]] m^2/s) \\ &=& [[return sf_latex(namespace_em_induction.emf)]] V \end{eqnarray} $$

Direction by right hand grib rule:

• $\mathcal{E} \gt 0$: counterclockwise.
• $\mathcal{E} \lt 0$: clockwise.

By Ohm's law, the magnitude of the current: $$ \begin{eqnarray} I &=& \frac{V}{R} \\ &=& \frac{|\mathcal{E}|}{R} \\ &=& \frac{[[return sf_latex(Math.abs(namespace_em_induction.emf))]] V}{~r~ \Omega} \\ &=& [[return sf_latex(Math.abs(namespace_em_induction.current))]] A \end{eqnarray} $$ The direction of $\mathcal{E}$ earlier gives the direction of the current.

The magnitude of the force: $$ \begin{eqnarray} F &=& |I \vec L \times \vec B| \\ &=& |([[return sf_latex(Math.abs(namespace_em_induction.current))]] A) (~l~ m) ([[return Math.abs(~b~)]] T)| \\ &=& [[return sf_latex(Math.abs(namespace_em_induction.force))]] N \end{eqnarray} $$

The direction of the force can be found with the right hand rule for cross product. However, by Lenz's law, we know that the force is always opposite to the change, so it must point opposite to the direction of the velocity $ \vec v$. Since $\vec v$ points [[return (~v~>0)? "right":"left"]], $\vec F$ must points [[return (~v~>0)? "left":"right"]].

Solution

Magnitude of the magnetic force on the charge: $$ \begin{eqnarray} F_B = |q \vec v \times \vec B| \\ &=& ([[return Math.abs(~q~)]] C) ([[return Math.abs(~v~)]]m/s) ([[return Math.abs(~b~)]] T) \\ &=& [[return sf_latex(Math.abs(namespace_em_induction.force))]]N \end{eqnarray} $$ The force points [[return (namespace_em_induction.force>0)? "up":"down"]], as can be seen by applying the right hand rule.

From the point of view of a charge on the bar, the magnetic force can be interpreted as an electric force: $\vec F_B = q \vec v \times \vec B \equiv q \vec E_{eff}$. This gives: $$ \begin{eqnarray} \vec E_{eff} &=& \vec v \times \vec B \\ \Rightarrow E_{eff} &=& |\vec v \times \vec B| & \text{ writing $E_{eff} = |\vec E_{eff}|$} \\ &=& ([[return Math.abs(~v~)]]m/s) ([[return Math.abs(~b~)]] T) \\ &=& [[return sf_latex(Math.abs(namespace_em_induction.e_field))]] V/m \end{eqnarray} $$ The effective E field points [[return (namespace_em_induction.e_field>0)? "up":"down"]], as can be seen by applying the right hand rule.

Remember $E_y = -\frac{dV}{dy} \Rightarrow dV = -E_y dy$. The magnitude of the potential difference is: $$ \begin{eqnarray} |\Delta V| &=& E_{eff} |\Delta y| \\ &=& (E_{eff}) (l)\\ &=& ([[return sf_latex(Math.abs(namespace_em_induction.e_field))]] V/m) (~l~ m) \\ &=& [[return sf_latex(Math.abs(namespace_em_induction.emf))]] V \end{eqnarray} $$ This potential difference is in fact the same as the emf $\mathcal{E} = -\frac{d\Phi_B}{dt}$ we computed in the earlier problem. This is just an alternative way of understanding it.

The electric force produced by $E_{eff}$ is: $$ \begin{eqnarray} \vec F_E &=& q \vec E_{eff} \\ \Rightarrow F_E &=& |q \vec E_{eff}| \\ &=& ([[return Math.abs(~q~)]] C) ([[return sf_latex(Math.abs(namespace_em_induction.e_field))]] N/C) & \text{ remember $V/m = N/C$} \\ &=& [[return sf_latex(Math.abs(namespace_em_induction.force))]]N \end{eqnarray} $$ This force (both direction and magnitude) must be identical to the magnetic force $F_B$ you calcalated earlier because they are just different ways to view the same force.

To cancel the force above, we must impose $E_{opposite}$ to point opposite to $E_{eff}$, i.e. $\vec E_{opposite} = - \vec E_{eff}$. Therefore, the direction of $E_{eff}$ before tells us $E_{opposite}$ points [[return (namespace_em_induction.e_field>0)? "down":"up"]].

Alternatively, we can think of $E_{opposite}$ generating an electric force that cancels the effect of the magnetic field: $$ \begin{eqnarray} \vec 0 &=& q \vec v \times \vec B + q \vec E_{opposite} \\ \Rightarrow q \vec E_{opposite} &=& -q \vec v \times \vec B \\ \vec E_{opposite} &=& -\vec v \times \vec B = -\vec E_{eff} \end{eqnarray} $$ This leads to the same result.

Solution

$s_{left}$

The distance between the long wire and the left side is: $$ \begin{eqnarray} s_{left} &=& s [[return (~sign_x_wire~ > 0)? "+": "-"]] \frac{w}{2} \\ &=& ~x_wire~ m [[return (~sign_x_wire~ > 0)? "+": "-"]] \frac{~width_loop~ m}{2} \\ &=& [[return sf_latex(s_left)]] m \end{eqnarray} $$

$s_{right}$

The distance between the long wire and the left side is: $$ \begin{eqnarray} s_{right} &=& s [[return (~sign_x_wire~ > 0)? "-": "+"]] \frac{w}{2} \\ &=& ~x_wire~ m [[return (~sign_x_wire~ > 0)? "-": "+"]] \frac{~width_loop~ m}{2} \\ &=& [[return sf_latex(s_right)]] m \end{eqnarray} $$

$\Phi$

In this solution, assume the $x$-axis points right, and the $z$-axis comes out of the screen. Since we assume $\vec A$ points out of the screen, we can write $d\vec A = dA \hat k$. $$ \begin{eqnarray} \Phi &=& \int \vec B \cdot d\vec A \\ &=& \int \vec B \cdot d(A \hat k) \\ &=& \int B_z dA \qquad \text{$B_z = \vec B \cdot \hat k$ is the component out of the screen.} \\ &=& \int\int B_z dx dy \\ &=& \int_{s-w/2}^{s+w/2} \left( [[return (sign_b_z > 0)? "+":"-"]]\frac{\mu_0 I_1}{2 \pi x} \right) dx \int_0^h dy \qquad \text{The sign of $B_z$ is decided by the right hand grip rule.} \\ &=& [[return (sign_b_z > 0)? "+":"-"]]\frac{\mu_0 I_1 h}{2 \pi} \int_{s-w/2}^{s+w/2} \frac{dx}{x} \\ &=& [[return (sign_b_z > 0)? "+":"-"]]\frac{\mu_0 I_1 h}{2 \pi} (\ln x)\bigg|_{s-w/2}^{s+w/2} \\ &=& [[return (sign_b_z > 0)? "+":"-"]]\frac{\mu_0 I_1 h}{2 \pi} \ln \frac{s + w/2}{s-w/2} \\ &=& [[return (sign_b_z > 0)? "+":"-"]]\frac{(4\pi \times 10^{-7}) (~current_wire~) (~height_loop~)}{2 \pi} \ln \frac{[[return sf_latex(s_max)]]}{[[return sf_latex(s_min)]]} \\ &=& [[return sf_latex(flux)]] Wb \end{eqnarray} $$

${\cal E}$

By Faraday's law: $$ \begin{eqnarray} {\cal E} &=& - \frac{d\Phi}{dt} \\ &=& -\frac{d}{dt} \bigg( [[return (sign_b_z > 0)? "+":"-"]]\frac{\mu_0 I_1 h}{2 \pi} \ln \frac{s + w/2}{s-w/2} \bigg) \\ &=& [[return (sign_b_z > 0)? "-":"+"]]\frac{\mu_0 h}{2 \pi} \ln \frac{s + w/2}{s-w/2} \frac{dI_1}{dt} \end{eqnarray} $$ Since $I_1$ is [[return (~sign_rate_change_current_wire~ > 0)? "increasing" : "decreasing"]], we have $\frac{dI_1}{dt} = [[return (~sign_rate_change_current_wire~ > 0)? "+" : "-"]] ~rate_change_current_wire~ A/s$. This gives: $$ \begin{eqnarray} {\cal E} &=& [[return (sign_b_z > 0)? "-":"+"]]\frac{\mu_0 h}{2 \pi} \ln \frac{s + w/2}{s-w/2} \frac{dI_1}{dt} \\ &=& [[return (sign_b_z > 0)? "-":"+"]]\frac{(4\pi \times 10^{-7}) (~height_loop~)}{2 \pi} \ln \frac{[[return sf_latex(s_max)]]}{[[return sf_latex(s_min)]]} ([[return (~sign_rate_change_current_wire~ > 0)? "+" : "-"]] ~rate_change_current_wire~) \\ &=& [[return sf_latex(emf)]] V \end{eqnarray} $$

Direction of $\cal E$

We defined $\vec A$ to point out of the screen. The right hand grip rules means you point your thumb out of the screen, then the four fingers define the positive direction as counterclockwise. Since we found $\cal E$ to be [[return (emf > 0)? "positive" : "negative"]], we know the direction of the induced emf and current to be [[return (emf > 0)? "counterclockwise" : "clockwise"]]. You can also apply Lenz's law to determine the direction.

Direction of forces

Once you figured out the direction of the induced current, you can apply $\vec F = I \vec L \times \vec B$ to deduce the forces. The forces are shown in the figure.

Direction of $F_{total}$

The forces on the top and bottom sides of the loop always cancel out by symmetry. The forces on the vertical sides always points in opposite direction, but the force on the side nearer to the long wire will always be stronger, this determines the direction of the total force.

Force on the long wire

By Newton's third law, the force on the long wire must be equal in magnitude to the force on the loop but opposite in direction.

Solution

Lenz's law states that the induced emf due to a change in a magnetic flux is directed to oppose the change.

Step	Answer	Explanation
Direction of $\vec B_{ext}$	[[return (sign_b_z>0)? "out of screen":"into screen"]]	$\vec B_{ext}$ is found by applying the right hand grip rule on the long wire.
Change in the strength of $\vec B_{ext}$	[[return (~sign_rate_change_current_wire~ > 0)? "increasing" : "decreasing"]]	Current in long wire [[return (~sign_rate_change_current_wire~ > 0)? "increasing" : "decreasing"]]
Direction of $\vec B_{induced}$	[[return (sign_b_z * ~sign_rate_change_current_wire~ >0)? "into screen": "out of screen"]]	$\vec B_{induced}$ opposes the change in $\vec B_{ext}$ (see * below for details)
Direction of $\mathcal{E}$	[[return (emf > 0)? "CCW": "CW"]]	Apply right hand grip rule to $\vec B_{induced}$

*: For step 3:
• If change (in step 2) is increasing then $\vec B_{ext}$ and $\vec B_{induced}$ are anti-parallel.
• If change (in step 2) is decreasing then $\vec B_{ext}$ and $\vec B_{induced}$ are parallel.

Direction of $\cal E$

We defined $\vec A$ to point out of the screen. The right hand grip rules means you point your thumb out of the screen, then the four fingers define the positive direction as counterclockwise. Since we found $\cal E$ to be [[return (emf > 0)? "positive" : "negative"]], we know the direction of the induced emf and current to be [[return (emf > 0)? "counterclockwise" : "clockwise"]]. You can also apply Lenz's law to determine the direction.

Direction of forces

Once you figured out the direction of the induced current, you can apply $\vec F = I \vec L \times \vec B$ to deduce the forces. The forces are shown in the figure.

Direction of $F_{total}$

The forces on the top and bottom sides of the loop always cancel out by symmetry. The forces on the vertical sides always points in opposite direction, but the force on the side nearer to the long wire will always be stronger, this determines the direction of the total force.

Force on the long wire

By Newton's third law, the force on the long wire must be equal in magnitude to the force on the loop but opposite in direction.