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?

Given the potential difference (\(V\)) that the sphere is accelerated through, we can find its final kinetic energy using the formula: \(E_k = qV\), where \(E_k\) is the kinetic energy and \(q\) is the charge of the sphere. The kinetic energy can then be used to determine the final velocity (\(v_f\)), using the formula: \(\frac{1}{2}mv^2 = E_k\), where \(m\) is the mass of the sphere. \(E_k = (5.00 \cdot 10^{-4} \mathrm{C})(7000 \mathrm{V}) = 3.50 \mathrm{J}\) Now, we can use the kinetic energy to calculate the final velocity: \(\frac{1}{2}(3.00 \cdot 10^{-6} \mathrm{kg})v^2 = 3.50 \mathrm{J}\) Solving for \(v\): \(v = \sqrt{\frac{2(3.50 \mathrm{J})}{(3.00 \cdot 10^{-6} \mathrm{kg})}} \approx 48391.48 \mathrm{m/s}\)

Now that we have the final velocity of the sphere, we can find the centripetal force acting on the sphere due to the magnetic field. The formula for this force is \(F_c = qvB \sin{\theta}\), where \(F_c\) is the centripetal force, \(B\) is the magnitude of the magnetic field, and \(\theta\) is the angle between the magnetic field and the velocity. In this case, the angle is \(90^\circ\), so we can simplify the formula to \(F_c = qvB\). \(F_c = (5.00 \cdot 10^{-4} \mathrm{C})(48391.48 \mathrm{m/s})(4.00 \mathrm{T}) \approx 96.78 \mathrm{N}\)

To find the radius of curvature (\(r\)) of the sphere's motion, we can use the centripetal force formula: \(F_c = \frac{mv^2}{r}\), where \(m\) and \(v\) are the mass and velocity of the sphere, respectively. Now, we can rearrange the formula to solve for \(r\). \(r = \frac{mv^2}{F_c}\) \(r = \frac{(3.00 \cdot 10^{-6} \mathrm{kg})(48391.48 \mathrm{m/s})^2}{96.78 \mathrm{N}}\) \(\approx 0.301 \mathrm{m}\) The radius of curvature of the sphere's motion in the magnetic field is approximately \(0.301 \mathrm{m}\).