Question

A long, straight wire is located along the \(x\) -axis \((y=0\) and \(z=0\) ). The wire carries a current of \(7.00 \mathrm{~A}\) in the positive \(x\) -direction. What are the magnitude and the direction of the force on a particle with a charge of \(9.00 \mathrm{C}\) located at \((+1.00 \mathrm{~m},+2.00 \mathrm{~m}, 0),\) when it has a velocity of \(3000 . \mathrm{m} / \mathrm{s}\) in each of the following directions? a) the positive \(x\) -direction b) the positive \(y\) -direction c) the negative \(z\) -direction

Short answer

Answer: The magnitude of the force is \(0 N\) and there is no force acting on the particle.

Step-by-step solution

We will use the Biot-Savart law to determine the magnetic field at the location of the particle. Since the current is along the positive \(x\) -axis and the particle is in the \(xy\) plane, the magnetic field vector will be perpendicular to the \(xy\) plane and pointing in the positive \(z\) direction. We can use the formula for the magnetic field due to an infinitely long straight wire: \(B=\frac{\mu_0 I}{2\pi r}\) Where \(B\) is the magnetic field, \(\mu_0\) is the permeability of free space (\(4\pi \times 10^{-7} \mathrm{Tm/A}\)), \(I\) is the current in the wire, and \(r\) is the distance from the wire to the particle. In our case, \(I = 7.00 A\) and \(r = 2.00 m\), so we can plug these values into the formula: \(B=\frac{(4\pi \times 10^{-7} \mathrm{Tm/A})(7.00 A)}{2\pi(2.00 m)}\) Calculating the magnetic field, we get: \(B=7.00 \times 10^{-5} T\) #Step 2: Calculate the Lorentz force on the charged particle#

Now that we have found the magnetic field at the location of the charged particle, we can use the Lorentz force equation to determine the force acting on the particle for each specified velocity direction. The Lorentz force equation is: \(\vec{F} = q(\vec{v} \times \vec{B})\) Where \(\vec{F}\) is the force vector, \(q\) is the charge of the particle, \(\vec{v}\) is its velocity vector, and \(\vec{B}\) is the magnetic field vector. For this problem, \(q = 9.00 C\), and we have to calculate the force for three different velocity directions: positive \(x\)-direction, positive \(y\)-direction, and negative \(z\)-direction. (a) Velocity in the positive \(x\)-direction

Calculate force for the velocity in the positive \(x\)-direction#

When the velocity is in the positive \(x\)-direction, the velocity vector is \(\vec{v} = (3000 m/s, 0, 0)\) Since the magnetic field vector is in the positive \(z\) direction, \(\vec{B} = (0, 0, 7.00 \times 10^{-5} T)\). Now we can calculate the cross product: \(\vec{F} = q(\vec{v} \times \vec{B}) = (0, -9.00 C(3000 m/s)(7.00 \times 10^{-5}T) , 0)\) Calculating the force, we get: \(\vec{F} = (0, -1.89 N, 0)\) So, the magnitude of the force is \(1.89 N\) acting in the negative \(y\)-direction. (b) Velocity in the positive \(y\)-direction

Calculate force for the velocity in the positive \(y\)-direction#

When the velocity is in the positive \(y\)-direction, the velocity vector is \(\vec{v} = (0, 3000 m/s, 0)\) Since the magnetic field vector is in the positive \(z\) direction, \(\vec{B} = (0, 0, 7.00 \times 10^{-5} T)\). Now we can calculate the cross product: \(\vec{F} = q(\vec{v} \times \vec{B}) = (9.00 C(3000 m/s)(7.00 \times 10^{-5}T) , 0 , 0) \) Calculating the force, we get: \(\vec{F} = (1.89 N, 0, 0)\) So, the magnitude of the force is \(1.89 N\) acting in the positive \(x\)-direction. (c) Velocity in the negative \(z\)-direction

Calculate force for the velocity in the negative \(z\)-direction#

When the velocity is in the negative \(z\)-direction, the velocity vector is \(\vec{v} = (0, 0, -3000 m/s)\) Since the magnetic field vector is in the positive \(z\) direction, \(\vec{B} = (0, 0, 7.00 \times 10^{-5} T)\). Now we can calculate the cross product: \(\vec{F} = q(\vec{v} \times \vec{B}) = (0, 0, 0)\) In this case, there is no force acting on the particle. Summary of the force magnitudes and directions: (a) Magnitude: \(1.89 N\) , Direction: negative \(y\)-direction (b) Magnitude: \(1.89 N\) , Direction: positive \(x\)-direction (c) Magnitude: \(0 N\) , Direction: no force acting on the particle