Answer:
38.7Ā°
Explanation:
The force on a current carrying wire is given as
F = (B)(I)(L) sin Īø
where
B = magnetic field = 80 mT = 0.08 T
I = current in the wire = 40 A
(F/L) = 2.0 N/m
Īø = angle between the magnetic field and the direction of current flow (basically angle between the wire and the magnetic field) = ?
(F/L) = (B) (I) sin Īø
2 = 0.08 Ć 40 Ć sin Īø
sin Īø = (2/3.2) = 0.625
Īø = sinā»Ā¹ (0.625)
Īø = 38.7Ā°
Hope this Helps!!!
Answer:
38.68 Ā°C
Explanation:
Using,
F = BILsinŠ¤....................... Equation 1
F/L = BIsinŠ¤...................... Equation 2
Where F/L = force per unit length, B = Magnetic Field, I = current, Š¤ = angle between the wire and the magnetic Field
make Š¤ the subject of the equation
Š¤ = sinā»Ā¹[(F/L)/BI]................... Equation 3
Given: F/L = 2 N/m, B = 80 mT = 0.08 T, I = 40 A.
Substitute into equation 3
Š¤ = sinā»Ā¹(2/(0.08Ć40))
Š¤ = sinā»Ā¹(0.625)
Š¤ = 38.68 Ā°C
Hence the angle between the wire and the magnetic Field = 38.68 Ā°C
