Answer:
5.9 x 10⁻⁶ T
Explanation:
The direction of magnetic field is clockwise around a long current carrying wire. Its value is given by the relation
B = Kx 2 current / distance [ K is a constant having value 10⁻⁷ for air.]
At the middle point between two given wires , magnetic field due to each of the wires will be equal and opposite.So they will cancel out each other to zero. Hence resultant magnetic field at this point will be zero.
Part C : At this point , magnetic field due to each of the wires will be added up because they act in the same direction.
Magnetic field due to first wire
B₁ = [tex]\frac{10^{-7}\times2\times4.8}{28\times10^{-2}}[/tex]
= 3.4 x 10⁻⁶ T
Magnetic field due to second wire
B₂ = [tex]\frac{10^{-7}\times2\times4.8}{38\times10^{-2}}[/tex]
= 2.5 X 10⁻⁶ T
Total field = ( 3.4 +2.5)x 10⁻⁶
= 5.9 x 10⁻⁶ T.
Its direction will be downwards on the plane of the paper.
(a) The magnetic field midway between the wires will point into the paper and the magnitude is 1.92 x 10⁻⁵T.
(b) The magnetic field at point 28 cm to the right of P1 will point out of the paper and the magnitude is 2.91 x 10⁻⁶T.
The magnetic field between the two parallel wires is calculated as follows;
[tex]B = \frac{\mu_o I}{2\pi d}[/tex]
where;
The magnetic field will point into the paper.
d = ¹/₂ x 10 cm = 5 cm
[tex]B = \frac{(4\pi \times 10^{-7}) \times (4.8)}{2\pi \times (0.05)} \\\\B = 1.92 \times 10^{-5} \ T[/tex]
The magnetic field will point out of the paper.
d = 5 cm + 28 cm = 33 cm
[tex]B = \frac{(4\pi \times 10^{-7}) \times (4.8)}{2\pi \times (0.33)} \\\\B = 2.91 \times 10^{-6} \ T[/tex]
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