The figure shows a closed surface that has a line integral of the magnetic field of 1.00×10-6 Tm. If I1 = 19.0 A, I2 = 14.0 A and I4 = 25.0 A, what is the current flowing through I3? 30.8 A, into the page 30.8 A, out of the page 14.8 A, out of the page 14.8 A, into the page
Question
The figure shows a closed surface that has a line integral of the magnetic field of 1.00×10-6 Tm. If I1 = 19.0 A, I2 = 14.0 A and I4 = 25.0 A, what is the current flowing through I3? 30.8 A, into the page 30.8 A, out of the page 14.8 A, out of the page 14.8 A, into the page
Solution
To solve this problem, we need to use Ampere's Law, which states that the integral of the magnetic field around a closed loop is equal to the current enclosed by the loop.
The line integral of the magnetic field is given as 1.00×10^-6 Tm. This is equal to the total current enclosed by the surface, multiplied by the permeability of free space (μ0).
μ0 is a constant equal to 4π × 10^-7 Tm/A.
So, we can write Ampere's Law as:
∮B⋅dl = μ0 * I_enclosed
Rearranging for I_enclosed gives:
I_enclosed = ∮B⋅dl / μ0
Substituting the given values:
I_enclosed = 1.00×10^-6 Tm / 4π × 10^-7 Tm/A
I_enclosed = 0.795 A
This is the total current enclosed by the surface.
The problem also gives the currents I1, I2, and I4. The total current is the sum of these currents and I3, which we are trying to find:
I_enclosed = I1 + I2 + I3 + I4
Substituting the given values:
0.795 A = 19.0 A + 14.0 A + I3 + 25.0 A
Solving for I3 gives:
I3 = 0.795 A - 19.0 A - 14.0 A - 25.0 A
I3 = -57.205 A
However, current cannot be negative. The negative sign indicates that the direction of I3 is opposite to the direction we assumed when applying Ampere's Law.
So, the current flowing through I3 is 57.205 A, into the page.
However, none of the given options match this answer. There may be a mistake in the problem or the given options.
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