Question

when the distance from the magnet doubles from 3 cm to 6 cm, the magnetic field becomes as strong: less than half, half, two times, more than two times

Explanation:

Step1: Recall the inverse - square law for magnetic (and electric) forces

The strength of magnetic (and electric) forces between two objects (like magnets or charged objects) follows an inverse - square relationship with the distance \(r\) between them. The general formula for the force \(F\) is \(F\propto\frac{1}{r^{2}}\), where \(r\) is the distance between the two objects.

Step2: Analyze the change in distance

Initially, the distance is \(r_1 = 3\space cm\). After doubling, the new distance is \(r_2=6\space cm\). We can find the ratio of the new force \(F_2\) to the initial force \(F_1\) using the inverse - square law.

From \(F\propto\frac{1}{r^{2}}\), we have \(F_1\propto\frac{1}{r_1^{2}}\) and \(F_2\propto\frac{1}{r_2^{2}}\). Then \(\frac{F_2}{F_1}=\frac{\frac{1}{r_2^{2}}}{\frac{1}{r_1^{2}}}=\frac{r_1^{2}}{r_2^{2}}\)

Substitute \(r_1 = 3\space cm\) and \(r_2 = 6\space cm\) into the formula: \(\frac{F_2}{F_1}=\frac{3^{2}}{6^{2}}=\frac{9}{36}=\frac{1}{4}\)? Wait, no, wait. Wait, the question is about how the magnetic field (force) becomes as strong. Wait, maybe I misread. Wait, the initial distance is \(3\space cm\), then it doubles to \(6\space cm\). The force is inversely proportional to the square of the distance. So if \(r\) becomes \(2r\) (since \(6 = 2\times3\)), then \(F\) becomes \(F'=\frac{1}{(2r)^{2}}=\frac{1}{4r^{2}}\) while the original force \(F=\frac{1}{r^{2}}\). But wait, the options are about how much of the original strength. Wait, maybe the question is phrased as "the magnetic field becomes ____ as strong". Let's re - evaluate.

Wait, the inverse - square law: \(F\propto\frac{1}{r^{2}}\). Let the initial distance be \(d\), force \(F_1\propto\frac{1}{d^{2}}\). When distance is \(2d\), force \(F_2\propto\frac{1}{(2d)^{2}}=\frac{1}{4d^{2}}\). But wait, the options are "half", "less than half", "two times", "more than two times". Wait, maybe I made a mistake. Wait, no, if \(r\) doubles, the force is \(\frac{1}{4}\) of the original? But that's not one of the options. Wait, maybe the question is about the magnetic field strength around a magnet, and the relationship is approximately inverse - square, but maybe in some cases, for the region near a magnet, but the key is that when distance increases, the force (field strength) decreases.

Wait, initial distance \(r_1 = 3\space cm\), final distance \(r_2=6\space cm\). The factor by which distance increases is \(k = \frac{r_2}{r_1}=2\). The force is inversely proportional to the square of the distance, so the force is inversely proportional to \(k^{2}=4\). So the new force is \(\frac{1}{4}\) of the original? But \(\frac{1}{4}\) is less than \(\frac{1}{2}\) (since \(\frac{1}{4}=0.25\) and \(\frac{1}{2} = 0.5\)). So the magnetic field (force) becomes less than half as strong.

Answer:

less than half