the images below show four pairs of magnets. the magnets in different pairs do not affect each other. all…

Explanation:

Step1: Determine the relationship between magnetic force and distance

The magnetic force between two magnets is inversely proportional to the square of the distance between them. That is, (F\propto\frac{1}{d^{2}}). So, the closer the magnets, the stronger the magnetic force.

Step2: Compare the distances in each pair

• Pair 1 has a distance of (1.5) inches.
• Pair 2 has a distance of (3.5) inches.
• Pair 3 has a distance of (1.5) inches.
• Pair 4 has a distance of (3.5) inches.

Since Pair 1 and Pair 3 have smaller distances ((d = 1.5) inches) compared to Pair 2 and Pair 4 ((d=3.5) inches). Now, consider the orientation: in Pair 1, the poles are (N - S) (attractive), in Pair 3, the poles are (S - S) (repulsive). But the magnitude of the magnetic force (attractive or repulsive) depends on the distance. Since the distance in Pair 1 and Pair 3 is the same, but we assume that when calculating the magnitude of the force (ignoring the direction - attractive or repulsive), the formula (F = k\frac{m_1m_2}{d^{2}}) (where (m_1) and (m_2) are the magnetic moments of the magnets). If we assume the magnets in Pair 1 have larger magnetic moments (because one magnet is larger) compared to Pair 3 (where magnets are of the same size as the smaller magnet in Pair 1).

Let’s assume the magnetic moment is proportional to the size of the magnet (for magnets of the same material). Let (m_{1 - large}) be the magnetic moment of the large magnet in Pair 1, (m_{1 - small}) be the magnetic moment of the small magnet in Pair 1, and (m_{3}) be the magnetic moment of magnets in Pair 3 (equal to (m_{1 - small})).

The force in Pair 1 (F_1=k\frac{m_{1 - large}\times m_{1 - small}}{(1.5)^{2}}), and the force in Pair 3 (F_3 = k\frac{m_{3}\times m_{3}}{(1.5)^{2}}). Since (m_{1 - large}>m_{1 - small}=m_{3}), then (F_1>F_3)

Answer:

Pair 1