Magnetics 5 Question 9
9. Two short bar magnets of length $1 cm$ each have magnetic moments $1.20 Am^{2}$ and $1.00 Am^{2}$, 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 $20.0 cm$. 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 the earth’s magnetic induction is $3.6 \times 10^{-5} Wb / m^{2}$ )
(2013 Main)
(a) $3.6 \times 10^{-5} Wb / m^{2}$
(b) $2.56 \times 10^{-4} Wb / m^{2}$
(c) $3.50 \times 10^{-4} Wb / m^{2}$
(d) $5.80 \times 10^{-4} Wb / m^{2}$
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Answer:
Correct Answer: 9. (b)
Solution:
- $B _{\text {net }}=B _1+B _2+B _H$
$$ \begin{aligned} B _{net} & =\frac{\mu _0}{4 \pi} \frac{\left(M _1+M _2\right)}{r^{3}}+B _H \\ & =\frac{10^{-7}(1.2+1)}{(0.1)^{3}}+3.6 \times 10^{-5} \\ & =2.56 \times 10^{-4} Wb / m^{2} \end{aligned} $$
