Complete question is;
One long wire carries a current of 30 A along the entire x axis. A second long wire carries a current of 40 A perpendicular to the xy plane and passes through the point (0, 4, 0) m. What is the magnitude of the resulting magnetic field at the point y = 2.0 m on the y axis?
Answer:
B_net = 50 × 10^(-7) T
Explanation:
We are told that the 30 A wire lies on the x-plane while the 40 A wire is perpendicular to the xy plane and passes through the point (0,4,0).
This means that the second wire is 4 m in length on the positive y-axis.
Now, we are told to find the magnitude of the resulting magnetic field at the point y = 2.0 m on the y axis.
This means that the position we want to find is half the length of the second wire.
Thus, at this point the net magnetic field is given by;
B_net = √[(B1)² + (B2)²]
Where B1 is the magnetic field due to the first wire and B2 is the magnetic field due to the second wire.
Now, formula for magnetic field due to very long wire is;
B = (μ_o•I)/(2πR)
Thus;
B1 = (μ_o•I_1)/(2πR_1)
Also, B2 = (μ_o•I_2)/(2πR_2)
Now, putting the equation of B1 and B2 into the B_net equation, we have;
B_net = √[((μ_o•I_1)/(2πR_1))² + ((μ_o•I_2)/(2πR_2))²]
Now, factorizing out some common terms, we have;
B_net = (μ_o/2π)√[((I_1)/R_1))² + ((I_2)/R_2))²]
Now,
μ_o is a constant and has a value of 4π × 10^(−7) H/m
I_1 = 30 A
I_2 = 40 A
Now, as earlier stated, the point we are looking for is 2 metres each from wire 2 end and wire 1.
Thus;
R_1 = 2 m
R_2 = 2 m
So, let's calculate B_net.
B_net = ((4π × 10^(−7))/2π)√[(30/2)² + (40/2)²]
B_net = 50 × 10^(-7) T
is its wavelength
and v=1400 m/s, therefore its frequency is
