Two long bar 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 on the other.

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Solution

Here $M_{1} = M_{2}$ = 10 A.m

and r= 2 cm = 0.02 m

We know that,

Force exterted by two magnetic poles on each other

= $\frac{μ_{0}}{4 π} \frac{M_{1} M_{2}}{r^{2}}$

= $\frac{4 π \times 10^{- 7} \times 10^{2}}{4 π \times 4 \times 10^{- 4}}$

= $2.5 \times 10^{- 2}$ N.

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Two long bar 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 on the other.

Here M1=M2 = 10 A.m and r= 2 cm = 0.02 m We know that, Force exterted by two magnetic poles on each other = μ04πM1M2r2 = 4π×10−7×1024π×4×10−4 = 2.5×10−2N.

Here $M_{1} = M_{2}$ = 10 A.m

and r= 2 cm = 0.02 m

We know that,

Force exterted by two magnetic poles on each other

= $\frac{μ_{0}}{4 π} \frac{M_{1} M_{2}}{r^{2}}$

= $\frac{4 π \times 10^{- 7} \times 10^{2}}{4 π \times 4 \times 10^{- 4}}$

= $2.5 \times 10^{- 2}$ N.