Question

Two short bar magnets of length 1 cm each have magnetic moments $1.20A{m}^2$ and $1.00A{m}^2$ respectively. They are placed on a horizontal table parallel to each other with their $N$ poles pointing towards the South. They have a common magnetic equator and are separated by a distance of $20.0cm$. The value of the resultant horizontal magnetic induction at the mid-point $O$ of the line joining their centres is close to
(Horizontal component of earth's magnetic induction is $3.6\times {10}^{-5}Wb/m^2$)

• A. $2.56\times {10}^{-4}Wb/m^2$
• B. $3.50\times {10}^{-4}Wb/m^2$
• C. $5.80\times {10}^{-4}Wb/m^2$
• D. $3.6\times {10}^{-5}Wb/m^2$

Answer 1

The correct option is A $2.56\times {10}^{-4}Wb/m^2$

Given that,

$m_1=1.20A.m^2$

$m_2=1.00A.m^3$

B along equator :

$B=\frac{{\mu }_o}{4\pi }\frac{M}{d^3}$

Here $B_1=\frac{{\mu }_0}{4\pi }\times \frac{1.20}{{10}^{-3}}$

$B_1={10}^{-4}\times 1.20Wb/m^2$

$B_2=\frac{{\mu }_0}{4\pi }\times \frac{1}{{10}^{-3}}={10}^{-4}Wb/m^2$

$\theta =0^{\circ }$

$\therefore Bresultant=10$:

$B=\sqrt{B_1^2+B_2^2+2B_1{B}_2}$ we know resultant will be directed from $NrightarrowS$ thus $\theta =0^{\circ }$

$=\left(\sqrt{1.44+1+1.4}\right)\times {10}^{-4}$

$=\sqrt{4.94}\times {10}^{-4}$

$B=2.226\times {10}^{-4}Wb/m^2$\Rightarrow a$

Thus $B=2.56\times {10}^{-4}Wb/m^2$