Question

Two long bare magnets are placed with their axes coinciding in such a way that the north pole of the first magnet is 2.0 cm from the south pole of the second. If both the magnets have a pole strength of 10 Am, find the force exerted by one magnet of the other.

Answer 1

Explanation:

To find: The force exerted by one magnet of the other

Step 1:

Given data in the question:

Pole strength = m1 = m2 = 10 A-m

Range between first magnet's north pole and second magnet's south pole,

r = 2 cm = 0.02 m

Step 2:

We know that,

Force F exerted on one another by two magnetic poles is given by

$F=\frac{\mu_{0}}{4 \pi}=\frac{m_{1} m_{2}}{r^{2}}$

$=\frac{\left(4 \pi \times 10^{-7} \times 10^{2}\right)}{4 \pi \times 4 \times 10^{-4}}$

$=\frac{\left(10^{-7} \times 10^{2}\right)}{4 \times 10^{-4}}$

$=\frac{10^{-5}}{4 \times 10^{-4}}$

$=\frac{10^{-1}}{4}$

$\begin{aligned}&=0.25 \times 10^{-1}\\&=2.5 \times 10^{-2} N\end{aligned}$