Question

the images below show four pairs of magnets. the magnets in different pairs do not affect each other. all the magnets shown are made of the same material, but some of them are different sizes and shapes. think about the magnetic force between the magnets in each pair. select the pair with the magnetic force of smallest magnitude. pair 1: image of a small bar magnet (s-n) and a circular magnet (n-s) with distance 0.55 cm pair 2: image of a circular magnet (s-n) and a small bar magnet (n-s) with distance 0.85 cm pair 3: image of a longer bar magnet (s-n) and a circular magnet (n-s) with distance 0.55 cm pair 4: image of a small bar magnet (n-s) and a circular magnet (s-n) with distance 0.85 cm

Explanation:

Step1: Recall Magnetic Force Factors

Magnetic force magnitude between magnets depends on distance (inverse - related, greater distance → smaller force) and pole strength (related to magnet size/shape, larger magnets or more aligned poles might have stronger force, but here same material; also, attraction/repulsion magnitude depends on distance and pole interaction).

Step2: Analyze Distance and Magnet Size

- Pair 1: Distance = 0.55 cm, small bar magnet.
- Pair 2: Distance = 0.85 cm, small bar magnet.
- Pair 3: Distance = 0.55 cm, larger bar magnet (longer than Pair 1's bar magnet).
- Pair 4: Distance = 0.85 cm, larger bar magnet (longer than Pair 2's bar magnet).

First, compare distance: 0.85 cm is greater than 0.55 cm, so Pairs 2 and 4 have larger distance than 1 and 3. Now, between Pair 2 and Pair 4: Pair 2 has a small bar magnet, Pair 4 has a larger bar magnet. Larger magnets (same material) have more magnetic material, so their pole strength is higher. So, for same distance (0.85 cm), Pair 2 has smaller bar magnet (less pole strength) than Pair 4. Also, distance is larger (0.85 cm) than 0.55 cm, and magnet size in Pair 2 is small. Now, check interaction: all pairs have opposite poles (attraction) or same? Wait, Pair 1: left bar S - N, right circle N - S → opposite poles (attraction). Pair 2: left circle S - N, right bar N - S → opposite poles (attraction). Pair 3: left bar S - N, right circle N - S → opposite poles (attraction). Pair 4: left bar N - S, right circle S - N → opposite poles (attraction). So all are attractive. Now, magnetic force magnitude: depends on distance (inverse square law - like Coulomb's law for magnetism, $F \propto \frac{1}{r^2}$) and pole strength (proportional to magnet size, since same material). So:

- For distance: larger distance (0.85 cm) gives smaller force than 0.55 cm. So between Pairs 2,4 (0.85 cm) and 1,3 (0.55 cm), Pairs 2 and 4 are candidates for smaller force.

- Between Pair 2 and Pair 4: Pair 2 has a small bar magnet, Pair 4 has a larger bar magnet. So Pair 2 has less pole strength than Pair 4. So with same distance (0.85 cm), Pair 2 has smaller pole strength, so smaller force. Also, compare Pair 2 with others: Pair 2 has distance 0.85 cm (largest distance among all) and small magnet (smallest pole strength among those with 0.85 cm distance). So Pair 2 should have the smallest magnetic force magnitude. Wait, wait, let's re - check:

Wait, Pair 2: distance 0.85 cm, bar magnet is small (same as Pair 1's bar magnet). Pair 4: distance 0.85 cm, bar magnet is large (same as Pair 3's bar magnet). So pole strength: Pair 4's bar magnet is larger, so higher pole strength than Pair 2's. So at same distance (0.85 cm), Pair 2 has smaller pole strength, so smaller force. Now, compare Pair 2 with Pair 1: Pair 1 has distance 0.55 cm (smaller than 0.85 cm), so even with small bar magnet, the smaller distance might give higher force than Pair 2's larger distance. Let's confirm: magnetic force $F \propto \frac{m_1m_2}{r^2}$, where $m_1, m_2$ are pole strengths, $r$ is distance.

For Pair 1: $r = 0.55$, $m_1$ (small bar), $m_2$ (circle magnet).

For Pair 2: $r = 0.85$, $m_1$ (circle magnet), $m_2$ (small bar).

For Pair 3: $r = 0.55$, $m_1$ (large bar), $m_2$ (circle magnet).

For Pair 4: $r = 0.85$, $m_1$ (large bar), $m_2$ (circle magnet).

Since $r$ in Pair 2 is 0.85 (larger than 0.55), and $m_1, m_2$ in Pair 2 are small (compared to Pair 4's large bar), the product $m_1m_2$ in Pair 2 is smaller than in Pair 4, and $r^2$ is larger. So $F$ for Pair 2 is smaller than Pair 4. And $F$ for Pair 2 is smaller than Pair 1 (because $r$ is larger in Pair 2, and even if $m_1m_2$ is similar, $r^2$ is $(0.85/0.55)^2\approx 2.37$, so force would be about 1/2.37 times, and if $m_1m_2$ is same, but Pair 2's bar is small, so $m_2$ (bar) in Pair 2 is small, so $m_1m_2$ in Pair 2 is smaller than Pair 1's $m_1m_2$ (since Pair 1's bar is small, same as Pair 2's bar? Wait, Pair 1 and Pair 2's bar magnets: looking at the images, Pair 1's bar is small (same as Pair 2's bar), Pair 3 and Pair 4's bars are large. So Pair 1: small bar (m) and circle (M). Pair 2: circle (M) and small bar (m). Pair 3: large bar (M') and circle (M). Pair 4: large bar (M') and circle (M).

So $F_1 \propto \frac{mM}{(0.55)^2}$, $F_2 \propto \frac{mM}{(0.85)^2}$, $F_3 \propto \frac{M'M}{(0.55)^2}$, $F_4 \propto \frac{M'M}{(0.85)^2}$.

Since $(0.85)^2 > (0.55)^2$, so $F_2 < F_1$ and $F_4 < F_3$. Also, $M' > m$, so $F_4 > F_2$ (because $M'M > mM$ and denominator is same for $F_2$ and $F_4$? Wait no, $F_2$ denominator is $(0.85)^2$, $F_4$ denominator is $(0.85)^2$. Numerator for $F_2$: $mM$, numerator for $F_4$: $M'M$. Since $M' > m$, $M'M > mM$, so $F_4 > F_2$. And $F_1$: numerator $mM$, denominator $(0.55)^2$. $F_2$: numerator $mM$, denominator $(0.85)^2$. Since $(0.85)^2 > (0.55)^2$, $F_2 < F_1$. Similarly, $F_3 > F_1$ (since $M' > m$), $F_4 > F_2$. So the smallest force is Pair 2.

Answer:

Pair 2