Question

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

• A. $3.60\times {10}^{-5}Wb/m^2$
• B. $2.56\times {10}^{-4}Wb/m^2$
• C. $3.50\times {10}^{-4}Wb/m^2$
• D. $5.80\times {10}^{-4}Wb/m^2$

Answer 1

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

$B_1=\frac{{\mu }_o}{4\pi }\frac{M_1}{{(d^1+l^2)}^{3/2}}{B}_2=\frac{{\mu }_o}{4\pi }\frac{M_2}{{(d^1+l^2)}^{3/2}}{B}_1={10}^{-7}\times \frac{1.2}{{({0.1}^2+{\left(0.505\right)}^2)}^{3/2}}=1.1955\times {10}^{-4}wb/㎡B_2={10}^{-7}\times \frac{1}{{({0.1}^2+{\left(0.505\right)}^2)}^{3/2}}=0.9963\times {10}^{-4}{B}_E=Earthmagneticfield{B}_E=0.36\times {10}^{-4}wb/m^2{B}_1+B_2+B_E=2.5518\times {10}^{-4}wb/m^2$