Question

Initially at rest, a small copper sphere with a mass of \(3.00 \cdot 10^{-6} \mathrm{~kg}\) and a charge of \(5.00 \cdot 10^{-4} \mathrm{C}\) is accelerated through a \(7000 .-\mathrm{V}\) potential difference before entering a magnetic field of magnitude \(4.00 \mathrm{~T}\), directed perpendicular to its velocity. What is the radius of curvature of the sphere's motion in the magnetic field?

Short answer

Answer: The radius of curvature of the copper sphere's motion in the magnetic field is 0.350 meters.

Step-by-step solution

Calculate the final velocity of the copper sphere

First, we need to find the final velocity of the copper sphere after it is accelerated through the \(7000\) V potential difference. To do this, we will use the formula for the kinetic energy gained by the sphere: \(KE = \frac{1}{2}m v^2 = qV\) where \(KE\) is the kinetic energy, \(m\) is the mass of the sphere, \(v\) is its final velocity, \(q\) is its charge, and \(V\) is the potential difference. Plugging in the given values, we have: \((3.00 \cdot 10^{-6}\mathrm{~kg}) \cdot v^2 = 2 \cdot (5.00 \cdot 10^{-4}\mathrm{C}) \cdot (7000\mathrm{~V})\) Now, solving for \(v\), we get: \(v^2 = \frac{2 \cdot (5.00 \cdot 10^{-4}\mathrm{C}) \cdot (7000 \mathrm{~V})}{(3.00 \cdot 10^{-6} \mathrm{~kg})}\) \(v = \sqrt{\frac{2 \cdot (5.00 \cdot 10^{-4}\mathrm{C}) \cdot (7000\mathrm{~V})}{(3.00 \cdot 10^{-6}\mathrm{~kg})}}\)

Calculate the magnetic force acting on the sphere

Now that we have the final velocity of the sphere, we can find the magnetic force acting on it using the following formula: \(F_{m} = qvB\sin(\theta)\) Since the magnetic field is perpendicular to the sphere's velocity, \(\theta = 90^{\circ}\) and \(\sin(\theta)=1\). Plugging in the values, we get: \(F_{m} = (5.00 \cdot 10^{-4} \mathrm{C}) \cdot v \cdot (4.00 \mathrm{~T})\) Notice that we already have an expression for \(v\) from Step 1, so let's substitute it here: \(F_{m} = (5.00 \cdot 10^{-4} \mathrm{C}) \cdot \sqrt{\frac{2 \cdot (5.00 \cdot 10^{-4}\mathrm{C}) \cdot (7000\mathrm{~V})}{(3.00 \cdot 10^{-6}\mathrm{~kg})}} \cdot (4.00 \mathrm{~T})\)

Calculate the radius of curvature of the sphere's motion

The magnetic force is the centripetal force acting on the sphere, which is responsible for the circular motion. The centripetal force can be expressed as: \(F_{c} = \frac{mv^2}{r}\) Since \(F_{m} = F_{c}\), we can equate the two expressions and solve for the radius \(r\): \((5.00 \cdot 10^{-4} \mathrm{C}) \cdot \sqrt{\frac{2 \cdot (5.00 \cdot 10^{-4}\mathrm{C}) \cdot (7000\mathrm{~V})}{(3.00 \cdot 10^{-6}\mathrm{~kg})}} \cdot (4.00 \mathrm{~T}) = \frac{(3.00 \cdot 10^{-6} \mathrm{~kg}) \cdot \left(\sqrt{\frac{2 \cdot (5.00 \cdot 10^{-4}\mathrm{C}) \cdot (7000\mathrm{~V})}{(3.00 \cdot 10^{-6} \mathrm{~kg})}}\right)^2}{r}\) Now cancel out some terms and solve for \(r\): \(r = \frac{(3.00 \cdot 10^{-6} \mathrm{~kg}) \cdot \left(\frac{2 \cdot (5.00 \cdot 10^{-4}\mathrm{C}) \cdot (7000\mathrm{~V})}{(3.00 \cdot 10^{-6} \mathrm{~kg})}\right)}{(5.00 \cdot 10^{-4} \mathrm{C}) \cdot( 4.00 \mathrm{~T})}\) \(r = 0.350\mathrm{~m}\) So the radius of curvature of the sphere's motion in the magnetic field is \(0.350\) meters.