Problem: The slinky is a coil of wire with about M turns. The slinky can be stretched to different lengths. In this problem, we will model the magnetic field of the slinky (if it carries current) using the infinite solenoid approximation. The slinky has a radius of about “Cm. (a) Draw a side view of the slinky, clearly indicating the current direction. Draw the magnetic field for your assumed current direction in the infinite solenoid approximation. Suppose an infinite straight wire passes through the point in the middle of the solenoid. Draw the direction the current must flow through this wire for the magnetic force to be directed toward the top of the page. (b) Compute the magnetic field for a slightly stretched slinky (length 2 cm) and an extremely stretched slinky (length M). Assume the slinky carries a current of IA. (c) Compute the self-inductance of the slightly stretched slinky (length = 2 km) and the extremely stretched slinky (length = “). Assume the slinky carries a current of AA. (d) Compute the magnetic energy stored in the slightly stretched slinky (length = ” cm) and an extremely stretched slinky (length = %). Assume the slinky carries a current of 44. (e) Identify the generic physical principle that a system will change to lower its energy. Will the slinky try to contract or expand?
The Correct Answer and Explanation is:
Let’s break down the problem step-by-step:
(a) Magnetic Field and Current Direction
- Side view of the slinky: A slinky is made of coils, and when it carries a current, each loop creates a magnetic field. The current direction through the slinky is assumed to be moving in the wire loops along the circumference of each coil. If you view the slinky from the side, the direction of the current through each loop follows the right-hand rule. For instance, if you curl the fingers of your right hand in the direction of the current in the slinky, your thumb points in the direction of the magnetic field inside the slinky.
- Magnetic Field in Infinite Solenoid Approximation:
- Inside the slinky, the magnetic field is uniform and parallel to the axis of the coil. It is similar to the field in an infinite solenoid, where the field lines are straight and point in the direction of the solenoid’s axis. Outside the solenoid (or slinky), the magnetic field is nearly zero, assuming the length of the slinky is large compared to its radius.
- Direction of Current in the Infinite Wire:
If we assume the infinite straight wire passes through the center of the slinky and you want the magnetic force on this wire to point upward (toward the top of the page), the current in the wire must flow perpendicular to both the wire and the magnetic field. Applying the right-hand rule for the force on a current-carrying wire in a magnetic field (i.e., F = I * (L × B)), the current in the infinite wire should flow to the left or into the page. This ensures the magnetic force on the wire will be directed upward.
(b) Magnetic Field for Different Slinky Lengths
For a solenoid (or slinky), the magnetic field inside is given by: B=μ0NLIB = \mu_0 \frac{N}{L} IB=μ0LNI
Where:
- μ0\mu_0μ0 is the permeability of free space.
- NNN is the number of turns of the slinky.
- LLL is the length of the slinky.
- III is the current flowing through the slinky.
- Slightly stretched slinky (length = 2 cm): For a slightly stretched slinky, the field will be stronger because the number of coils per unit length is larger.
- Extremely stretched slinky (length = M): As the slinky gets stretched, the number of coils per unit length decreases, reducing the magnetic field inside the slinky.
(c) Self-Inductance
Self-inductance LLL for a solenoid is given by the formula: L=μ0N2ALL = \mu_0 \frac{N^2 A}{L}L=μ0LN2A
Where:
- AAA is the cross-sectional area of the solenoid (which can be calculated as πr2\pi r^2πr2, with rrr being the radius of the slinky).
- Slightly stretched slinky (length = 2 cm): The inductance will be higher due to the smaller length, which increases the coil density and thus the inductance.
- Extremely stretched slinky (length = M): The inductance will be lower due to the larger length and lower coil density.
(d) Magnetic Energy Stored
The magnetic energy stored in a solenoid is given by the formula: U=12LI2U = \frac{1}{2} L I^2U=21LI2
Where:
- LLL is the inductance, and
- III is the current.
- Slightly stretched slinky (length = 2 cm): The energy stored will be greater since the inductance is higher for a more tightly wound coil.
- Extremely stretched slinky (length = M): The energy stored will be lower due to the reduced inductance.
(e) Physical Principle – System Will Change to Lower Its Energy
The system will seek to minimize its energy. The energy stored in the magnetic field depends on the inductance, which in turn depends on the length of the slinky. If the slinky is stretched, it stores less energy because of the lower inductance. Therefore, the slinky will tend to contract to reduce the length, increasing its inductance and lowering the overall energy stored in the system.
Conclusion
In summary:
- (a) The magnetic field inside a slinky is similar to that of an infinite solenoid, and the current in an external wire must flow to the left or into the page for the force to be directed upward.
- (b) The magnetic field inside a slinky will be stronger when it is shorter and weaker when it is stretched.
- (c) The self-inductance is higher for the shorter slinky and lower for the stretched one.
- (d) The magnetic energy stored is higher in the shorter slinky.
- (e) The slinky will contract to minimize its energy.
