Induced Currents and Magnetic Forces
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AP Physics C: Electricity and Magnetism › Induced Currents and Magnetic Forces
A rectangular wire loop is pulled at a constant speed $$v$$ out of a region of uniform magnetic field $$B$$. This results in a magnetic braking force of magnitude $$F_0$$. If the experiment is repeated with a magnetic field of strength $$2B$$, but all other parameters (loop dimensions, resistance, speed) remain the same, what will be the new magnetic braking force?
$$ F_0 / 2 $$
$$ 2F_0 $$
$$ 4F_0 $$
$$ F_0 / 4 $$
Explanation
The magnetic braking force is given by the expression $$ F = \frac{B^2 L^2 v}{R} $$. This equation shows that the force is directly proportional to the square of the magnetic field strength ($$ F \propto B^2 $$). Therefore, if the magnetic field strength $$B$$ is doubled, the force will increase by a factor of $$ 2^2 = 4 $$. The new force will be $$ 4F_0 $$.
A square conducting loop is moved at a constant velocity $$ \vec{v} $$ from a region of no magnetic field into a region with a uniform magnetic field $$ \vec{B} $$ directed into the page. The velocity vector is perpendicular to the boundary of the field and to one side of the loop. What is the direction of the magnetic force on the loop as it enters the field?
In the direction opposite to $$ \vec{v} $$, acting to decelerate the loop.
Into the page, parallel to the direction of $$ \vec{B} $$.
In the direction of $$ \vec{v} $$, acting to accelerate the loop.
Out of the page, opposite to the direction of $$ \vec{B} $$.
Explanation
As the loop enters the field, the magnetic flux into the page increases. By Lenz's law, the induced current creates a magnetic field out of the page to oppose this change. The right-hand rule for current loops indicates this requires a counter-clockwise current. The leading edge of the loop (the one inside the field) has an upward current. Using the Lorentz force right-hand rule ($$ \vec{F} = I\vec{L} \times \vec{B} $$), the force on this leading edge is to the left, opposite to the direction of velocity $$ \vec{v} $$. The forces on the top and bottom segments cancel each other out.
A square loop of wire is pulled at a constant speed $$v$$ first into a uniform magnetic field region, and then later pulled out of the same region at the same constant speed $$v$$. Let $$ F_{in} $$ be the magnitude of the magnetic force on the loop as it enters, and $$ F_{out} $$ be the magnitude as it exits. How do $$ F_{in} $$ and $$ F_{out} $$ compare?
$$ F_{in} = F_{out} $$ because the magnitude of the rate of change of flux is the same in both cases.
$$ F_{in} > F_{out} $$ because the direction of induced current weakens the interaction when exiting.
$$ F_{in} < F_{out} $$ because exiting the field requires more work against the induced field.
$$ F_{out} = 0 $$ while $$ F_{in} \neq 0 $$ because induction only occurs when flux increases.
Explanation
The magnitude of the induced emf is given by Faraday's Law, $$ |\mathcal{E}| = |-\frac{d\Phi_B}{dt}| $$. As the loop enters and exits at the same speed $$v$$, the rate at which the loop's area within the field changes is the same ($$dA/dt = Lv$$). Therefore, the magnitude of the rate of change of flux, $$ |d\Phi_B/dt| = B(Lv) $$, is identical for both entering and exiting. This means the magnitude of the induced emf and current $$I = |\mathcal{E}|/R$$ are also the same. Since the magnetic force magnitude is $$F = ILB$$, the forces $$F_{in}$$ and $$F_{out}$$ must be equal in magnitude.
A conducting ring is dropped from rest so that it falls through a finite region of uniform magnetic field directed perpendicular to its plane. The force of gravity acts on the ring. Which statement best describes the ring's downward acceleration?
The acceleration is less than $$g$$ as it enters, equal to $$g$$ while inside, and greater than $$g$$ as it exits.
The acceleration is less than $$g$$ as it enters and exits the field, and equal to $$g$$ while fully inside.
The acceleration is constant and equal to $$g$$ throughout its fall.
The acceleration is always less than $$g$$ as long as any part of the ring is in the field region.
Explanation
As the ring enters the field, the changing magnetic flux induces a current, which results in an upward magnetic braking force opposing gravity. The net downward force is less than $$mg$$, so the downward acceleration is less than $$g$$. While the ring is fully inside the uniform field, the magnetic flux is constant, so no current is induced and no magnetic force acts. The only force is gravity, so the acceleration is $$g$$. As the ring exits, the flux changes again, inducing another upward braking force, and the acceleration is again less than $$g$$.
A rectangular loop of wire enters a region of uniform magnetic field at a constant speed $$v$$, resulting in a magnetic braking force $$F$$. If the experiment is repeated with a new loop made of a wire with twice the resistivity but otherwise identical dimensions, moving at the same speed $$v$$, what will be the new braking force?
$$ 2F $$
$$ F $$
$$ F/2 $$
$$ F/4 $$
Explanation
The magnetic braking force is given by $$ F = \frac{B^2 L^2 v}{R} $$. The resistance $$R$$ of a wire is directly proportional to its resistivity $$ \rho $$ ($$R = \rho \frac{\text{length}}{\text{area}}$$). If the resistivity doubles, the total resistance $$R$$ of the loop also doubles. Since the force $$F$$ is inversely proportional to the resistance $$R$$, doubling the resistance will cause the braking force to be halved. The new force will be $$F/2$$.
A square loop of wire is moving at a constant speed $$v$$ through a non-uniform magnetic field $$B(x)$$ that is directed perpendicular to the loop and varies only with position $$x$$. At a particular instant, the leading edge is at $$x_2$$ and the trailing edge is at $$x_1$$. The existence of a net magnetic force on the loop is a direct consequence of which of the following conditions?
The speed of the loop is constant throughout its motion.
The magnetic field is different at the locations of the leading and trailing edges.
The resistance of the loop is finite and non-zero.
The average magnetic field over the loop's area is non-zero.
Explanation
A net magnetic force arises from two effects: an induced current and the interaction of that current with the field. In this scenario, a net motional EMF is induced only if the EMF on the leading edge ($$\mathcal{E}_2 = B(x_2)Lv$$) is different from the EMF on the trailing edge ($$\mathcal{E}_1 = B(x_1)Lv$$). This difference, which drives the current, exists only if $$B(x_2) \neq B(x_1)$$. Furthermore, the forces on these edges ($$F_2=ILB(x_2)$$ and $$F_1=ILB(x_1)$$) will only be unbalanced if the field strengths are different. Therefore, the fundamental condition for a net force is that the magnetic field is different at the two edges.
After a long time, the rod slides down the rails at a constant terminal velocity $$ v_T $$. What is the expression for this terminal velocity?
$$ v_T = \frac{mgR \sin\theta}{B^2 L^2} $$
$$ v_T = \frac{mgR}{B^2 L^2 \sin\theta} $$
$$ v_T = \frac{mgR \sin\theta}{B^2 L^2 \cos^2\theta} $$
$$ v_T = \frac{mgR \tan\theta}{B^2 L^2} $$
Explanation
At terminal velocity, the component of the gravitational force along the rails ($$mg \sin\theta$$) is balanced by the component of the magnetic braking force opposing it. The magnetic flux perpendicular to the loop is $$ \Phi_B = (B \cos\theta)(Lx) $$, where $$x$$ is the distance along the rails. The induced emf is $$ \mathcal{E} = |d\Phi_B/dt| = BLv\cos\theta $$. The current is $$ I = \mathcal{E}/R $$. The magnetic force on the rod is $$ F_B = ILB $$, directed horizontally. The component of this force up the incline is $$ F_{B, \parallel} = F_B \cos\theta = (ILB)\cos\theta = \frac{(BLv\cos\theta)LB\cos\theta}{R} = \frac{B^2 L^2 v \cos^2\theta}{R} $$. Setting $$ mg \sin\theta = F_{B, \parallel} $$ and solving for $$v=v_T$$ gives $$ v_T = \frac{mgR \sin\theta}{B^2 L^2 \cos^2\theta} $$.
A rectangular conducting loop is pushed at a constant velocity $$ \vec{v} $$ completely through and out of a finite region of uniform magnetic field $$ \vec{B} $$ directed out of the page. What is the direction of the net magnetic force on the loop as its leading edge exits the field, while its trailing edge is still inside?
In the direction of $$ \vec{v} $$.
There is no net magnetic force on the loop.
Opposite to the direction of $$ \vec{v} $$.
Out of the page, parallel to $$ \vec{B} $$.
Explanation
As the loop exits the field, the magnetic flux out of the page is decreasing. According to Lenz's law, the induced current will flow in a direction that creates a magnetic field out of the page to oppose this change. A right-hand rule analysis shows this requires a counter-clockwise current. The trailing edge of the loop is still inside the field and has an upward-directed current. Using the Lorentz force rule ($$ \vec{F} = I\vec{L} \times \vec{B} $$), an upward current in an out-of-page magnetic field results in a force directed opposite to the velocity. This is a braking force.
A conducting loop is given an initial velocity $$ \vec{v}_0 $$ to the right and enters a region of uniform magnetic field directed into the page. There are no other forces acting on the loop, such as gravity or friction. What happens to the loop's velocity as it moves through the field?
The velocity remains constant at $$ \vec{v}_0 $$ because the net magnetic force is always zero.
The velocity decreases as it enters and then increases back to $$ v_0 $$ as it exits due to energy conservation.
The velocity decreases continuously as long as any part of the loop is in the magnetic field.
The velocity decreases as it enters the field, becomes constant while fully inside, and decreases again as it exits.
Explanation
As the loop enters the field, a change in flux induces a current and a magnetic braking force, which opposes the motion and causes the loop to slow down. While fully inside the uniform field, the flux is constant, so there is no induced current and no magnetic force; the loop moves at a constant, reduced velocity. As the loop exits, the flux changes again, inducing another braking force that causes the loop to slow down further. The kinetic energy lost is dissipated as thermal energy in the loop.
A conducting rod of length $$L$$ is rotated with constant angular velocity $$ \omega $$ about a pivot at one of its ends, in a plane perpendicular to a uniform magnetic field $$ \vec{B} $$. This induces a current $$I$$ which flows radially along the rod (assuming a complete circuit). What is the resulting magnetic force on the rod?
A tangential force creating a torque that opposes the rotation.
A radial force directed toward the pivot, compressing the rod.
A tangential force creating a torque that assists the rotation.
A radial force directed away from the pivot, stretching the rod.
Explanation
Due to the rotation, a motional emf is induced, driving a current $$I$$ along the rod's length. The magnetic force on a current element $$I d\vec{r}$$ is $$ d\vec{F} = I d\vec{r} \times \vec{B} $$. With the current $$I$$ flowing radially and the field $$ \vec{B} $$ perpendicular to the plane of rotation, the force $$ d\vec{F} $$ is tangential. By Lenz's law, this force must oppose the motion that creates the current. Therefore, the force is directed tangentially opposite to the direction of motion, creating a magnetic torque that opposes the angular velocity.