Answer:
B = 5.32 x 10⁻⁵ T
Explanation:
given,
number of turn in the loop(N) = 22
radius of loop(R) = 13.5 cm = 0.135 m
current in the wire(I) = 0.52 A
μ₀ = 4 π x 10⁻⁷ H/m
magnitude of magnetic field (B)= ?
[tex]B = \dfrac{\mu_0 N I}{2R}\ \hat{R}[/tex]
[tex]B = \dfrac{4\pi \times 10^{-7}\times 22 \times 0.52}{2\times 0.135}[/tex]
[tex]B = \dfrac{1.4375\times 10^{-5}}{0.27}[/tex]
[tex]B = \dfrac{1.4375\times 10^{-5}}{0.27}[/tex]
B = 5.32 x 10⁻⁵ T
The magnitude of magnetic field at the center of loop is equal to B = 5.32 x 10⁻⁵ T
