# Magnitude of Current through a Rectangular Loop, given the Magnetic field

1. Mar 18, 2012

### blastoise

1. The problem statement, all variables and given/known data
A rectangular loop with dimensions 4.2 cm by 9.5 carries current I. The current in the loop produces a magnetic field at the center of the loop that has magnitude 5.60×10−5 T and direction away from you as you view the plane of the loop.

2. Relevant equations

$$\vec{B}=\frac{μ_0}{4\pi} \frac{Id\vec{l}\times \vec{r}}{r^2} (1.0) \ \ :|\vec{B}| =5.60 \times 10^{-5}; μ_0:= 4\pi \times 10^{-7} .$$

Where, dl is length, r hat is the unit vector, r^2 is source to point (would like to get this checked though not sure).
3. The attempt at a solution

Coordinate system: Holding up our left hand making an L with the index finger and thumb, let the thumb point in the + X direction index finger point in the +Y direction. Then the -Z direction will be defined in the direction we look.

Recall the unit vector is written in the form <i,j,k>. It becomes apparent using the "Right hand rule" the B field is pointing in the -k direction at all points; Thus, using the Principle of Superposition of Magnetic fields - The total magnetic field caused by several moving charges is the vector sum of the fields caused by the individual charges one can sum the magnetic field B created by the 4 wire segments(the dl vector is a vector with length dl, in the same direction as the current in the conductor).

Let B_1 be when L = 9.5 cm = .095m and r = 4.2cm /2 = .021 m is my r correct?
Then,
$$B_1=\frac{μ_0}{4\pi} \frac{I*L}{r^2} = \frac{μ_0}{4\pi} \frac{I*.095m}{.021^2m^2} .$$

Since there are two sides with 9.5 cm that create a B field in the -K direction the B field produced them is 2*B_1.

Let B_2 be when L = 4.5 cm = .045m and r = 9.5 cm / 2 = .0475m; Then,

$$B_2=\frac{μ_0}{4\pi} \frac{I*L}{r^2} = \frac{μ_0}{4\pi} \frac{I*.045m}{.0475^2m^2} .$$

Since there are two sides with 4.5cm that create a B field in the -K direction the B field produced by them is 2*B_2.

The Total B field is then 2B_1 + 2B_2

$$5.60×10^{-5}T=\frac{μ_0}{4\pi} I2[ \frac{.095 m}{.021^2m^2}+ \frac{.042m}{.0475^2m^2}] .$$

^ was wondering if that is correct.

(1.0) is Law of Biot and Savart.

Last edited: Mar 18, 2012