Question

The magnetic moment of a bar magnet is 0.256 $A{m}^2$. Its pole strength is 400 $mA$. It is cut into two equal pieces and these two pieces are arranged at right angles to each other with their unlike poles in contact (or like poles in contact). The resultant magnetic moment of the system is:

• A. $\sqrt{2}\times 256\times {10}^{-3}A{m}^2$
• B. $250\times {10}^{-3}A{m}^2$
• C. $\frac{256}{\sqrt{2}}\times {10}^{-3}A{m}^2$
• D. $\frac{128}{\sqrt{2}}\times {10}^{-3}A{m}^2$

Answer 1

Answer: (C)

$\frac{256}{\sqrt{2}}\times {10}^{-3}A{m}^2$

T$M_1=0.256A{m}^2$

$P_1=400milliAm$

$L_1=L$

$L_2=L/2,P_2=P_1$

$\frac{{\mu }_1}{L_1}=\frac{{\mu }_2}{L_2}$

$\frac{0.256}{L}=\frac{{\mu }_2}{L/2}$

${\mu }_2=0.128$

$=0.128\sqrt{2}A{m}^2$ (since angle is right angle so directly use vector sum property)

$=\frac{256}{\sqrt{2}}\times {10}^{-3}A{m}^2$