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What is the magnitude of the equatorial fields due to a bar magnet of length \( 5.0 cm \) at a distance \( 75 cm \) from its mid point? The magnetic moment of the bar magnet is \( 0.75 Am ^{2} \). (A) \( 3.2 \times 10^{-7} T \)(B) \( 1.78 \times 10^{-7} T \) (C) \( 6.4 \times 10^{-7} T \) (D) \( 3.56 \times 10^{-7} T \) |
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Answer» Correct option is (B) 1.76 x 10-7 T B = \(\frac{\mu_0}{4\pi}\frac{M}{(r^2+l^2)^{3/2}}\) B = \(\frac{10^{-7}\times0.75}{[(0.05)^2+(0.75)^2]^{3/2}}\) B = \(\frac{0.75\times10^{-7}}{[0.0025+0.5625]^{3/2}}\) B = \(\frac{0.75\times10^{-7}}{[0.565]^{3/2}}\) B = \(\frac{0.75\times10^{-7}}{0.424}\) B = 1.76 x 10-7 T |
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