A long straight wire carrying a 3.0 A current enters a room through a window 1.5 m high and 1.0 m wide. The path integral around the window frame has the value (in T.m):

Question

A long straight wire carrying a 3.0 A current enters a room through a window 1.5 m high and 1.0 m wide. The path integral  around the window frame has the value (in T.m):

Quizplus · Accepted Answer

Using Ampere's Law, we can find the magnetic field tangential to the window frame since it is a long straight wire:
$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}$
$I_{enc}$ is the current enclosed by the path. Since the current is entering the room through the window, the current enclosed by the path is simply 3.0 A. The path integral is the total length of the window frame, which is $1.5+1.0+1.5+1.0 = 5.0$ m.

So we have:
$BL = \mu_0 I_{enc}$
$B(5.0) = \mu_0 (3.0)$
$B = \frac{\mu_0 I_{enc}}{L} = \frac{4\pi	imes10^{-7}	imes3.0}{5.0}$ T

Now, the magnetic flux through the window is given by:
$\Phi_B = \int_S \vec{B} \cdot d\vec{A}$
Since the window is flat and perpendicular to the wire, the magnetic field is constant over the area of the window. Therefore, we can simply multiply the magnetic field by the area of the window to get the magnetic flux:
$\Phi_B = B\cdot A = \frac{4\pi	imes10^{-7}	imes3.0}{5.0} \cdot 1.5\cdot1.0 = 1.8	imes10^{-7}$ Wb.

The answer must be in Tesla-meters, so we need to multiply by the perimeter of the window to get the correct units:
$\Phi_B = B\cdot L = \frac{4\pi	imes10^{-7}	imes3.0}{5.0} \cdot 5.0 = 7.54	imes10^{-7}$ Tm.

This is closest to choice D, which is 3.8 $\mu$Tm (note that $\mu$T is the same as T).

A Long Straight Wire Carrying a 3