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I. Multiple Choice Questions · Q12

Q.Two short bar magnets have magnetic moments 1.20 A m2^2 and 1.00 A m2^2 respectively. They are kept on a horizontal table parallel to each other with their north poles pointing towards 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 (horizontal component of Earth's magnetic induction is 3.6×10−53.6\times10^{-5} Wb m−2^{-2}) (NSEP 2000-2001)

(a) 3.60×10−53.60\times10^{-5} Wb m−2^{-2}
(b) 3.5×10−53.5\times10^{-5} Wb m−2^{-2}
(c) 2.56×10−42.56\times10^{-4} Wb m−2^{-2}
(d) 2.2×10−42.2\times10^{-4} Wb m−2^{-2}
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Step 1. Both magnets lie flat on the table with a common magnetic equator and north poles pointing the same relative way ("towards south"), so the midpoint O sits on the equatorial line of each magnet individually, at r=10.0r=10.0 cm =0.10=0.10 m from each magnet's centre.

Step 2. Use the short-magnet equatorial field, Beq=μ04πpmr3B_{eq}=\dfrac{\mu_0}{4\pi}\dfrac{p_m}{r^3}, with μ0/4π=10−7\mu_0/4\pi=10^{-7}: for magnet 1 (pm=1.20p_m=1.20 A m2^2), B1=10−7×1.20/(0.10)3=10−7×1.20/0.001=1.2×10−4B_1 = 10^{-7}\times1.20/(0.10)^3 = 10^{-7}\times1.20/0.001 = 1.2\times10^{-4} T.

Step 3. For magnet 2 (pm=1.00p_m=1.00 A m2^2), B2=10−7×1.00/0.001=1.0×10−4B_2 = 10^{-7}\times1.00/0.001 = 1.0\times10^{-4} T. …

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