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(D) The horizontal component of the Earth's magnetic field is $H = 0.36 \; G$.At the null point on the axial line,the magnetic field of the magnet $B_

Vedclass · Accepted Answer

$0.54$

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(D) The horizontal component of the Earth's magnetic field is $H = 0.36 \; G$.At the null point on the axial line,the magnetic field of the magnet $B_{axial}$ is equal to the horizontal component of the Earth's magnetic field $H$:$B_{axial} = \frac{\mu_0}{4\pi} \frac{2M}{d^3} = H = 0.36 \; G \dots (i)$where $d = 14 \; cm$ is the distance from the center.The magnetic field on the equatorial line (normal bisector) at the same distance $d$ is given by:$B_{equatorial} = \frac{\mu_0}{4\pi} \frac{M}{d^3} = \frac{H}{2} = \frac{0.36}{2} = 0.18 \; G$.Since the magnetic field of the magnet on the equatorial line is in the same direction as the Earth's horizontal magnetic field,the total magnetic field $B_{total}$ at that point is:$B_{total} = B_{equatorial} + H = 0.18 \; G + 0.36 \; G = 0.54 \; G$.

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