A wire carrying a current is shaped in the form of a circular loop of radius 4.0 mm.If the $\vec { B }$ field magnitude at its center is $1.1 \mathrm { mT }$ with no external magnetic fields present,what is the magnitude of the current through the wire? (μ₀ = 4π × 10^-7 T ∙ m/A)

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The magnetic field at the center of a circular loop of wire is given by:$$B = \frac{\mu_0 I}{2R}$$where:* B is the magnetic field strength (in teslas)* μ0 is the vacuum permeability (4π × 10^-7 T ∙ m/A)* I is the current (in amperes)* R is the radius of the loop (in meters)In this problem, we are given:* B = 1.1 mT = 1.1 × 10^-3 T* R = 4.0 mm = 4.0 × 10^-3 mWe can solve for I by rearranging the equation:$$I = \frac{2BR}{\mu_0}$$Plugging in the given values, we get:$$I = \frac{2(1.1 	imes 10^{-3} 	ext{ T})(4.0 	imes 10^{-3} 	ext{ m})}{4\pi 	imes 10^{-7} 	ext{ T} \cdot 	ext{m/A}} = 7.0 	ext{ A}$$Therefore, the magnitude of the current through the wire is 7.0 A.

A Wire Carrying a Current Is Shaped in the Form B⃗\vec { B }B

A Wire Carrying a Current Is Shaped in the Form $\vec { B }$