Question

The force between two short Magets is F
When the pole strengths of the magnets are tripled and the distance between them is also doubled the force between them is (Take magnets to be Parallel
to each other)​

Answer 1

Given :

Force between two short magnets = F.

Pole strengths tripled.

Distance between them is doubled.

To find:

Force between magnets after changes.

Solution :

Magnetic force implies attraction or repulsion between charged particles due to its motion. Magnetic force is directly proportional to the pole strengths and inversely proportional to the distance between two poles .

Formula for magnetic force :

$\bf \: F = \frac{\mu_0 \times m_1 \times \: m_2 }{4\pi {r}^{2} }$

(equation 1)

where

μ0 = Permiability constant.

m1 = Pole strength of first magnet.

m2 = Pole strength of second magnet.

r = Distance between two magnets.

F = Magnetic force.

Let firstly,

Pole strengths of both magnets = m

Distance between two magnets = r

Putting values in equation 1,

Magnetic force in first case :

$\bf \: F_1 = \frac{\mu_0 \times m \times \: m }{4\pi {r}^{2} } \\ \\ \bf \: F_1 = \frac{\mu_0 \times {m}^{2} }{4\pi {r}^{2} }$

(equation 2)

In second case,

Pole strengths of both magnets is tripled,

So, pole strengths of both magnets = 3m

and Distance between both magnets is doubled.

So, Distance between both magnets = 2r.

Putting values in equation 1,

Magnetic force in second case :

$\bf \: F_2 = \frac{\mu_0 \times 3m \times \: 3m }{4\pi ({2r})^{2} } \\ \\ \bf \: F_2 = \frac{9(\mu_0 \times {m}^{2} )}{4(4\pi {r}^{2} )} \\ \bf \: \:$

(equation 3)

Putting the value of equation 2 in equation 3,

Magnetic force in second case :

$\bf \: F_2 \: = \frac{9}{4} F_1$

So, Magnetic force is increased 2.25 times than first case.