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A 6 cm long bar magnet possessing magnetic dipole moment 0.3 A-m^(2) is placed vertically on a horizontal wooden table. The north pole of the magnet touches the table. A neutral point is found on the table at a distance of 8 cm south of the magnet. Find the horizontal component of Earth's magnetic field. |
Answer» SOLUTION :Magnetic field due to single pole of a MAGNET is given by  `B = (mu_0)/(4pi) (M)/(d^(2))` where m is pole strength Field due to north pole , `B_N = (mu_0)/(4pi)(M)/((d^(2))) ` along NP Field due to south pole , `B_S = (mu_0)/(4pi) (M)/((d^(2) + 4l^(2))) ` along PS Horizontal component of `B_S = B_S cos theta ` `= (mu_0)/(4pi)(M)/((d^(2) + 4l^(2)))*(d)/((d^(2) + 4l^(2))^(1//2))` Resultant horizontal field ` = (mu_0 m)/(4pi) [ (1)/(d^(2)) -(d)/((d^(2) + 4l^(2))^(3//2))]` or , `B_H = (mu_0 M)/( 4pi (2l)) [ (1)/(d^(2)) - (d)/((d^(2) + 4l^(2))^(3//2))]` (where magnetic moment `M = mxx 2l` ) On SUBSTITUTING numerical values , we get `B_H = 3.81 xx 10^(-5)` Tesla. |
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