The force between two short bar magnet with magnetic moment M1 and M2, whose centers are r metre apart is 8N when their axes are in same line . If the separation is increased to 2r ,the force between them is reduced to?? please Explain (answer is 0.5 N)?

Question

Answer

In magnetic dipole force is directly proportional to 1÷r^4. here r=2 and 1 is replaced by 8N so; 8÷2^4----->8÷16--->0.5 thats it..hope u understand..!!👍

Answer

Magnetic force is inversally proportional to fourth power of distance between magnetic poles  .                     put the value and divid them . and ans is 0.5

Answer

The force between two short bar magnets can be determined using the formula:

$$F = \frac{{\mu_0}}{{4\pi}} \frac{{M_1M_2}}{{r^3}}$$

where F is the force between the magnets, $\mu_0$ is the permeability of free space, $M_1$ and $M_2$ are the magnetic moments of the magnets, and r is the distance between their centers.

Given that the force between the magnets is 8 N when their axes are in the same line, we can use this information to find the magnetic moments of the magnets.

$$\frac{{\mu_0}}{{4\pi}} \frac{{M_1M_2}}{{r^3}} = 8$$

Since the axes are in the same line, the magnets are aligned parallel to each other. In this case, the magnetic moments can be written as:

$M_1 = m_1 \cdot l_1$ and $M_2 = m_2 \cdot l_2$

where $m_1$ and $m_2$ are the magnetic pole strengths of the magnets, and $l_1$ and $l_2$ are the lengths of the magnets.

$$\frac{{\mu_0}}{{4\pi}} \frac{{(m_1 \cdot l_1)(m_2 \cdot l_2)}}{{r^3}} = 8$$

Simplifying the equation, we get:

$$\frac{{\mu_0}}{{4\pi}} \frac{{m_1m_2l_1l_2}}{{r^3}} = 8$$

Now, let's consider the situation where the separation between the magnets is increased to 2r. In this case, the new distance between their centers will be 2r.

Using the same formula, we can find the new force between the magnets:

$$F' = \frac{{\mu_0}}{{4\pi}} \frac{{m_1m_2l_1l_2}}{{(2r)^3}}$$

Simplifying this equation, we get:

$$F' = \frac{{\mu_0}}{{4\pi}} \frac{{m_1m_2l_1l_2}}{{8r^3}}$$

$$F' = \frac{1}{8} \left(\frac{{\mu_0}}{{4\pi}} \frac{{m_1m_2l_1l_2}}{{r^3}}\right)$$

Since we know that $F = 8$ N, we can substitute this value into the equation:

$$F' = \frac{1}{8} \cdot 8$$

$$F' = 1$$

Therefore, the force between the magnets when the separation is increased to 2r is 1 N, not 0.5 N as mentioned in the question.

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