Question

Force between two identical short bar magnets whose centres are $r$ metre apart is $4.8N$ when their axes are in the same line. If the separation is increased to $2r$ metre, the force between them is reduced to

• A. $0.3N$
• B. $0.6N$
• C. $24N$
• D. $1.2$

Answer 1

The correct option is A $0.3N$

Given

Two bar magnets placed axially at distance r

Force between them is 4.8 N

Solution

Force between two Magnetic dipoles is

$F=\frac{{\mu }_0\ast 6M_1{M}_2}{4\pi \ast d^4}$

F is inversely proportional to $r^4$

Therefore New Force will be

$F^{\prime }=\frac{F}{2^4}$

Because new d will be 2r

F'=4.8/16=0.3N

The correct answer is 0.3N