Question

Two short bar magnets have their magnetic moments $1.2A{m}^2$ and $1.0A{m}^2$. They are placed on a horizontal table parallel to each other at a distance of $20cm$ between their centres, such that their north poles pointing towards geographic south. They have common magnetic equatorial line. Horizontal component of earth's magnetic field is $3.6\times {10}^{-5}$ T. Then, the resultant horizontal magnetic induction at mid point of the line joining their centers is :
($\frac{{\mu }_o}{4\pi }={10}^{-7}H/m$)

• A. $3.6\times {10}^{-5}T$
• B. $1.84\times {10}^{-4}T$
• C. $2.56\times {10}^{-4}T$
• D. $5.8\times {10}^{-5}T$

Answer 1

The correct option is C $2.56\times {10}^{-4}T$

$B_{net}=B_{M_1}+B_{M_2}+B_H$

$=\frac{{\mu }_0{M}_1}{4\pi x^3}+\frac{{\mu }_0{M}_2}{4\pi x^3}+B_H$

$=\frac{{\mu }_0}{4\pi x^3}(M_1+M_2)+B_H$

$=\frac{{10}^{-7}}{{10}^{-3}}\times 2.2+3.6\times {10}^{-5}$ ; Here, $x={10}^{-1}m$

$=2.56\times {10}^{-4}T$