Magnetic Flux - AP Physics C Electricity and Magnetism

1. A flat, rectangular loop of wire is placed in a region of uniform magnetic field. For which orientation of the loop is the magnitude of the magnetic flux through it a maximum?
   Answer: The plane of the loop is perpendicular to the magnetic field. The magnetic flux is $\Phi_B = BA \cos\theta$, where $\theta$ is the angle between the field and the normal to the loop's plane. The magnitude is maximum when $|\cos\theta|$ is maximum, which occurs at $\theta = 0^\circ$ or $180^\circ$. In this case, the normal vector is parallel to the field, meaning the plane of the loop is perpendicular to the field.
2. A long straight wire carries a current $I$. A square loop of side length $a$ is placed a distance $a$ from the wire. The magnetic flux through the loop is calculated to be $\Phi_0$. If the current in the wire is doubled to $2I$ and the side length of the loop is halved to $a/2$ (while keeping the closest side at distance $a$), what is the new magnetic flux through the loop?
   Answer: Less than $\Phi_0$ The flux is $\Phi = \int B dA$. Initially, $\Phi_0 = \int_a^{2a} (\frac{\mu_0 I}{2\pi r}) (a \, dr) = \frac{\mu_0 I a}{2\pi} \ln(2)$. The new flux is $\Phi_{new} = \int_a^{a+a/2} (\frac{\mu_0 (2I)}{2\pi r}) (\frac{a}{2} \, dr) = \frac{\mu_0 I a}{2\pi} \int_a^{1.5a} \frac{dr}{r} = \frac{\mu_0 I a}{2\pi} \ln(1.5)$. Since $\ln(1.5) < \ln(2)$, the new flux is less than the original flux $\Phi_0$.
3. A square loop of wire with side length $s$ lies in the xy-plane with one corner at the origin and its sides aligned with the positive x and y axes. The loop is in a non-uniform magnetic field given by $\vec{B} = C y \hat{k}$, where $C$ is a positive constant. What is the total magnetic flux through the loop?
   Answer: $\frac{1}{2} C s^3$ The magnetic field varies with $y$. We must integrate to find the flux. The area element can be taken as a thin horizontal strip of width $s$ (in the x-direction) and height $dy$. So, $d\vec{A} = s \, dy \, \hat{k}$. The flux through this strip is $d\Phi_B = \vec{B} \cdot d\vec{A} = (C y \hat{k}) \cdot (s \, dy \, \hat{k}) = C s y \, dy$. Integrating from $y=0$ to $y=s$ gives $\Phi_B = \int_0^s C s y \, dy = C s [\frac{y^2}{2}]_0^s = \frac{1}{2} C s^3$.
4. A uniform magnetic field $\vec{B} = B_0 \hat{k}$ exists in space. What is the magnitude of the magnetic flux through the curved surface of a hemisphere of radius $R$ whose circular base lies in the xy-plane and is centered at the origin?
   Answer: $B_0 \pi R^2$ According to Gauss's law for magnetism, the total magnetic flux through any closed surface is zero. Consider the closed surface formed by the hemisphere and its flat circular base. $\Phi_{closed} = \Phi_{curved} + \Phi_{base} = 0$. Therefore, $\Phi_{curved} = -\Phi_{base}$. The flux through the flat base of area $A=\pi R^2$ in the xy-plane is $\Phi_{base} = \vec{B} \cdot \vec{A}_{base}$. With the outward normal for the base pointing in the $-z$ direction, $\vec{A}_{base} = -\pi R^2 \hat{k}$, so $\Phi_{base} = (B_0 \hat{k}) \cdot (-\pi R^2 \hat{k}) = -B_0 \pi R^2$. The flux through the curved surface is thus $\Phi_{curved} = -(-B_0 \pi R^2) = B_0 \pi R^2$.
5. The magnetic flux through a surface is calculated using the surface integral $\Phi_B = \int \vec{B} \cdot d\vec{A}$. This integral form, rather than the simpler $\Phi_B = \vec{B} \cdot \vec{A}$, is generally required under which of the following conditions?
   Answer: When either the magnetic field is non-uniform over the surface or the surface is not flat. The integral form is a general definition. It simplifies to the algebraic form $\Phi_B = \vec{B} \cdot \vec{A} = BA\cos\theta$ only if the magnetic field $\vec{B}$ is uniform over the entire surface and the surface is flat (planar). If either the field varies from point to point on the surface (is non-uniform) or the surface is curved, the integral is necessary to sum up the contributions of $d\Phi_B = \vec{B} \cdot d\vec{A}$ over all the infinitesimal area elements $d\vec{A}$ that make up the surface.
6. A square coil with $N$ turns and side length $s$ is in a uniform magnetic field $B$. The axis of the coil (normal to its plane) is initially parallel to the field. The coil is rotated by $90^\circ$ so its axis is perpendicular to the field. What is the magnitude of the magnetic flux through the entire coil in this final position?
   Answer: Zero The magnetic flux through the entire coil is $N$ times the flux through a single turn, $\Phi_{coil} = N(BA \cos\theta)$. Here, $A = s^2$. In the final position, the axis of the coil is perpendicular to the field, meaning the angle $\theta$ between the normal vector and the field is $90^\circ$. Since $\cos(90^\circ) = 0$, the magnetic flux through the coil is zero.
7. A rectangular loop of wire has initial width $w_0$ and length $L$. The loop is in a uniform magnetic field $B$ perpendicular to its plane. The loop is then stretched at a constant rate such that its width becomes $2w_0$ while its length remains $L$. What is the magnetic flux through the loop as a function of its instantaneous width $w$?
   Answer: $B L w$ The magnetic flux is given by $\Phi_B = BA$, where A is the instantaneous area of the loop, because the field is uniform and perpendicular to the loop's plane. The area of the rectangular loop at any instant is its length times its instantaneous width, $A = Lw$. Therefore, the magnetic flux as a function of the width $w$ is $\Phi_B = B(Lw) = BLw$.
8. Based on its fundamental definition involving magnetic field and area, the SI unit of magnetic flux, the weber (Wb), is equivalent to which of the following?
   Answer: Tesla-meter squared ($\text{T}\cdot\text{m}^2$) The fundamental definition of magnetic flux for a uniform field is $\Phi_B = BA \cos\theta$. The unit of magnetic field ($B$) is the tesla (T) and the unit of area ($A$) is the meter squared ($\text{m}^2$). Therefore, the unit of magnetic flux is tesla-meter squared ($\text{T}\cdot\text{m}^2$), which is defined as the weber (Wb). While a volt-second is also equivalent to a weber, it is derived from Faraday's law of induction, not the direct definition of flux.
9. A circular loop of wire with area $A$ is placed in a uniform magnetic field of strength $B$. What is the magnetic flux through the loop when the angle between the magnetic field vector and the normal to the plane of the loop is $60^\circ$?
   Answer: $\frac{1}{2} BA$ Magnetic flux is given by $\Phi_B = \vec{B} \cdot \vec{A} = BA \cos\theta$, where $\theta$ is the angle between the magnetic field vector and the area vector (which is normal to the loop's plane). With $\theta = 60^\circ$, $\cos(60^\circ) = 1/2$, so the flux is $\frac{1}{2} BA$.
10. The magnetic field perpendicular to the plane of a $4.0$ cm by $4.0$ cm square loop of wire is given by $B(t) = 3.0t - 1.5t^2$, where $B$ is in tesla and $t$ is in seconds. What is the magnitude of the magnetic flux through the loop at $t=2.0$ s?
    Answer: Zero First, calculate the area of the loop: $A = (0.040 \text{ m})^2 = 0.0016 \text{ m}^2$. Next, evaluate the magnetic field at $t=2.0$ s: $B(2.0) = 3.0(2.0) - 1.5(2.0)^2 = 6.0 - 1.5(4.0) = 6.0 - 6.0 = 0$ T. Since the magnetic field is zero at this instant, the magnetic flux $\Phi_B = BA$ is also zero.
11. What is the net magnetic flux through any closed surface, such as a sphere or a cube, regardless of the magnetic fields or currents present?
    Answer: Zero. This is a statement of Gauss's law for magnetism, $\oint \vec{B} \cdot d\vec{A} = 0$. Since there are no magnetic monopoles, magnetic field lines always form closed loops. Therefore, any field line that enters a closed surface must also exit it, resulting in a net magnetic flux of zero.
12. A circular loop is held in a uniform magnetic field. The flux through the loop is non-zero. The loop is then reshaped into a square of equal perimeter while remaining in the same field and orientation. How does the magnitude of the magnetic flux through the square compare to the original flux through the circle?
    Answer: The flux is less through the square. For a fixed perimeter, a circle encloses the maximum possible area. Let the perimeter be $P$. For the circle, $P=2\pi r$, so $r=P/(2\pi)$, and area $A_{circle}=\pi r^2 = P^2/(4\pi)$. For the square, side length is $s=P/4$, and area $A_{square}=s^2 = P^2/16$. Since $4\pi \approx 12.56 < 16$, we have $A_{circle} > A_{square}$. As flux is proportional to area ($\Phi=BA$), the flux through the square is less than the flux through the circle.
13. Which of the following correctly contrasts magnetic flux with electric flux?
    Answer: Magnetic flux through a closed surface must be zero, while electric flux through a closed surface is proportional to the enclosed charge. This statement correctly identifies the key difference embodied by Gauss's laws for electricity and magnetism. Due to the absence of magnetic monopoles, magnetic field lines are always closed loops, so the net magnetic flux through any closed surface is zero. Electric field lines, however, originate and terminate on charges, so the net electric flux through a closed surface is proportional to the net charge enclosed.
14. A long solenoid of radius $R$ has $n$ turns per unit length and carries a current $I$. The magnetic field inside is uniform and given by $B=\mu_0 n I$. A smaller, coaxial circular loop of radius $r < R$ is placed inside the solenoid. What is the magnetic flux through the smaller loop?
    Answer: $\mu_0 n I \pi r^2$ The magnetic field inside the solenoid is uniform with magnitude $B=\mu_0 n I$ and is directed along the axis. The flux is calculated through the area of the smaller loop, $A = \pi r^2$. Since the field is uniform and perpendicular to the plane of the inner loop, the flux is $\Phi_B = BA = (\mu_0 n I)(\pi r^2)$.
15. A long, straight wire carries a current $I$. A rectangular loop of wire with width $w$ and length $L$ is positioned such that its sides of length $L$ are parallel to the wire. The side closer to the wire is at a distance $a$ from it. The magnetic field due to the wire is $B = \frac{\mu_0 I}{2\pi r}$, where $r$ is the perpendicular distance from the wire. Which integral correctly gives the magnitude of the magnetic flux through the loop?
    Answer: $\int_a^{a+w} \frac{\mu_0 I L}{2\pi r} dr$ The magnetic field is not uniform over the area of the loop; it depends on the distance $r$ from the wire. To find the total flux, we must integrate. Consider a thin strip of the loop of length $L$ and width $dr$ at a distance $r$ from the wire. The area of this strip is $dA = L dr$. The flux through this strip is $d\Phi_B = B dA = (\frac{\mu_0 I}{2\pi r}) (L dr)$. To find the total flux, we integrate this expression over the width of the loop, from $r=a$ to $r=a+w$.
16. Which of the following statements provides the best conceptual definition of magnetic flux?
    Answer: A measure of the total number of magnetic field lines passing through a given surface area. Magnetic flux ($\Phi_B$) is conceptually a measure of the amount of magnetic field passing through a surface. It is quantified by counting the net number of magnetic field lines that pierce the surface.
17. A uniform magnetic field is given by the vector $\vec{B} = (4.0 \hat{i} - 3.0 \hat{j})$ T. A flat, square surface with side length $2.0$ m lies in the yz-plane. What is the magnetic flux through this surface?
    Answer: $16$ Wb The area of the square is $A = (2.0 \text{ m})^2 = 4.0 \text{ m}^2$. Since the surface lies in the yz-plane, its area vector is perpendicular to that plane, meaning it points in the x-direction. We can write $\vec{A} = 4.0 \hat{i} \text{ m}^2$. The magnetic flux is the dot product $\Phi_B = \vec{B} \cdot \vec{A} = ((4.0 \hat{i} - 3.0 \hat{j}) \text{ T}) \cdot (4.0 \hat{i} \text{ m}^2) = (4.0)(4.0) = 16$ Wb.
18. A flat surface of area $A$ is in a uniform magnetic field $B$. The magnetic flux through the surface is $\Phi_B = \frac{1}{2}BA$. What is the angle between the magnetic field vector and the plane of the surface?
    Answer: $30^\circ$ The magnetic flux is given by $\Phi_B = BA \cos\theta$, where $\theta$ is the angle between the magnetic field and the normal to the surface. We are given $\Phi_B = \frac{1}{2}BA$, so $BA \cos\theta = \frac{1}{2}BA$, which means $\cos\theta = 1/2$. This gives $\theta = 60^\circ$. This is the angle between the field and the normal. The question asks for the angle between the field and the plane of the surface, which is $90^\circ - \theta = 90^\circ - 60^\circ = 30^\circ$.
19. Loop 1 is a square with side length $2R$. Loop 2 is a circle with radius $R$. Both loops are in the same uniform magnetic field, and their planes are oriented perpendicular to the field lines. What is the ratio of the magnetic flux through Loop 1 to the magnetic flux through Loop 2, $\frac{\Phi_1}{\Phi_2}$?
    Answer: $\frac{4}{\pi}$ The magnetic flux is $\Phi = BA$, since the field is uniform and perpendicular to the planes. For Loop 1 (square), the area is $A_1 = (2R)^2 = 4R^2$. For Loop 2 (circle), the area is $A_2 = \pi R^2$. The ratio of the fluxes is $\frac{\Phi_1}{\Phi_2} = \frac{BA_1}{BA_2} = \frac{4R^2}{\pi R^2} = \frac{4}{\pi}$

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