Answer:
6.73 × 10⁻⁵
Explanation:
It is given that,
Number of turns, N = 23
Radius of the coil, r = 24.9 cm = 0.249 m
The distance coil axis 31.9 cm, x = 0.319m
Current flowing through the coils, I = 2.49 A
We need to find the magnitude of the magnetic field at a location on the axis of the coils
. The magnetic field of the coils is given by :
[tex]B=\dfrac{\mu_oINr^2}{(x^2+r^2)^{3/2}}B\\=\dfrac{4\pi \times 10^{-7}\times 2.49\times 23\times (0.249)^2}\\{(0.319^2+0.249^2)^{3/2}}B\\ = 6.73 \times 10^{-5}\\[/tex]
