Uniform current sheets are located in free space as follows: \(8 \mathbf{a}_{z} \mathrm{~A} / \mathrm{m}\) at \(y=0,-4 \mathbf{a}_{z} \mathrm{~A} / \mathrm{m}\) at \(y=1\), and \(-4 \mathbf{a}_{z} \mathrm{~A} / \mathrm{m}\) at \(y=-1\). Find the vector force per meter length exerted on a current filament carrying \(7 \mathrm{~mA}\) in the \(\mathbf{a}_{L}\) direction if the filament is located at \((a) x=0, y=0.5\), and \(\mathbf{a}_{L}=\mathbf{a}_{z} ;\) (b) \(y=0.5, z=0\), and \(\mathbf{a}_{L}=\mathbf{a}_{x} ;(c) x=0, y=1.5\), and \(\mathbf{a}_{L}=\mathbf{a}_{z}\)

First, we need to find the magnetic field produced by each current sheet. Since the current is flowing in the \(\mathbf{a}_z\) direction and the current sheets are on the \(y\) plane, the magnetic field will be in the \(\mathbf{a}_x\) direction. Using Amperè's law, we can find that: - Magnetic field due to the first current sheet at \(y=0\): \(\vec{B_1} = 4 \pi \times 8 \times 10^{-7} \mathbf{a}_x \mathrm{~T}\). For \(y<0\), \(\vec{B_1}\) is positive. For \(y>0\), \(\vec{B_1}\) is negative. - Magnetic field due to the second current sheet at \(y=1\): \(\vec{B_2} = 2 \pi \times 4 \times 10^{-7} \mathbf{a}_x \mathrm{~T}\), which is always positive. - Magnetic field due to the third current sheet at \(y=-1\): \(\vec{B_3} = 2 \pi \times 4 \times 10^{-7} \mathbf{a}_x \mathrm{~T}\), which is always negative.

For each case, we will now compute the total magnetic field at the location of the current filament by summing the magnetic fields produced by each current sheet: 1. Located at \((x, y, z) = (0, 0.5, 0)\): \(\vec{B} = -\vec{B_1} + \vec{B_2} = -2\pi \times 4 \times 10^{-7} \mathbf{a}_x \mathrm{~T}\). 2. Located at \((x, y, z) = (0, 0.5, 0)\): same as in the first case: \(\vec{B} = -2\pi \times 4 \times 10^{-7} \mathbf{a}_x \mathrm{~T}\). 3. Located at \((x, y, z) = (0, 1.5, 0)\): \(\vec{B} = -\vec{B_1} - \vec{B_2} = -6\pi \times 4 \times 10^{-7} \mathbf{a}_x \mathrm{~T}\).

Now, we need to find the force exerted on the current filament using the formula \(\vec{F} = \vec{I} \times \vec{B}\), where \(\vec{I}\) is the current in the filament and \(\vec{B}\) is the total magnetic field at the location of the filament: 1. Located in the \(\mathbf{a}_z\) direction: \(\vec{F} = (7 \times 10^{-3} \mathbf{a}_z) \times (-2\pi \times 4 \times 10^{-7} \mathbf{a}_x) = 0\). 2. Located in the \(\mathbf{a}_x\) direction: \(\vec{F} = (7 \times 10^{-3} \mathbf{a}_x) \times (-2\pi \times 4 \times 10^{-7} \mathbf{a}_x) = 0\). 3. Located in the \(\mathbf{a}_z\) direction: \(\vec{F} = (7 \times 10^{-3} \mathbf{a}_z) \times (-6\pi \times 4 \times 10^{-7} \mathbf{a}_x) = 0\). In all three cases, the vector force exerted on the current filament is zero. The reason is that the force on a current filament due to a magnetic field is perpendicular both to the magnetic field and to the direction of the current flow. In all cases, the magnetic field is either parallel or anti-parallel to the current flow direction, so the force is zero.