Answer:
Direction of current = clockwise
Magnitude of current, I = 0.36 A
Explanation:
The magnetic field strength, [tex]B_{E} = 50 \mu T[/tex]
The angle of dip, ∅ = 56°
The net magnetic field in the center of the tank is:
[tex]B_{net} = (B_{E} cos \phi ) (\hat{x} ) + ( B + B_{E} sin \phi)(-\hat{y})\\B_{net} = (50 cos 56 ) (\hat{x} ) + ( B +50 sin 56)(-\hat{y})\\B_{net} = (28 \mu T ) (\hat{x} ) + ( B +41.4 \mu T)(-\hat{y})\\[/tex]
The direction of the net magnetic field is:
[tex]\phi = tan^{-1} \frac{B + 41.4 }{28} \\tan \phi = \frac{B + 41.4 }{28}\\\phi = 62^0\\tan 62 = \frac{B + 41.4 }{28}\\28 tan 62 = B + 41.4\\52.66 = B + 41.4\\B = 11.26 \mu T[/tex]
The magnetic field due to the coil:
[tex]B = \frac{\mu_{0}NI }{2r} \\11.26 * 10^{-6} = \frac{4\pi * 10^{-7} * 50 *I }{2 *1}\\I = \frac{2 * 11.26 * 10^{-6}}{4\pi * 10^{-7} * 50} \\I = 0.36 A[/tex]
The current must be in clockwise direction to produce the field in downward direction