Respuesta :
Answer:
Magnetic field, [tex]B=2.39\times 10^{-3}\ T[/tex]
Explanation:
It is given that,
Number of turns, N = 320
Radius of the coil, r = 6 cm = 0.06 m
The distance from the center of one coil to the electron beam is 3 cm, x = 3 cm = 0.03 m
Current flowing through the coils, I = 0.5 A
We need to find the magnitude of the magnetic field at a location on the axis of the coils, midway between the coils. The magnetic field midway between the coils is given by :
[tex]B=\dfrac{\mu_oINr^2}{(x^2+r^2)^{3/2}}[/tex]
[tex]B=\dfrac{4\pi \times 10^{-7}\times 0.5\times 320\times (0.06)^2}{(0.03^2+0.06^2)^{3/2}}[/tex]
B = 0.00239 T
or
[tex]B=2.39\times 10^{-3}\ T[/tex]
So, the magnitude of the magnetic field at a location on the axis of the coils, midway between the coils is [tex]2.39\times 10^{-3}\ T[/tex]. Hence, this is the required solution.
