Question

A small aluminum ball with a charge of \(11.17 \mathrm{C}\) is moving northward at \(3131 \mathrm{~m} / \mathrm{s}\). You want the ball to travel in a horizontal circle with a radius of \(2.015 \mathrm{~m}\) and in a clockwise direction when viewed from above. Ignoring gravity, the magnitude of the magnetic field that must be applied to the aluminum ball to cause it to move in this way is \(B=0.8000 \mathrm{~T}\) What is the mass of the ball?

Short answer

Answer: The mass of the aluminum ball is approximately 0.03612 kg.

Step-by-step solution

Identify the relevant equations

In this case, two key equations are relevant - the magnetic force F_mag acting on the ball, and the centripetal force F_c required to keep the ball in a circular path. Magnetic force: \(F_{mag} = qvB\sin{\theta}\), where q is the charge, v is the velocity, B is the magnetic field, and \(\theta\) is the angle between the velocity and the magnetic field directions Centripetal force: \(F_c = \frac{mv^2}{r}\), where m is the mass, v is the velocity, and r is the radius of the circle. In this scenario, the magnetic force provides the centripetal force to keep the ball in a circular path. Therefore, we can set F_mag equal to F_c.

Set magnetic force equal to centripetal force

\(qvB\sin{\theta} = \frac{mv^2}{r}\) Given that the ball is moving in a horizontal circle (when viewed from above) and clockwise direction, the angle between the velocity and the magnetic field (\(\theta\)) will be 90 degrees (\(\sin{90°} = 1\)) for the required motion. We know the values of charge, velocity, radius, and magnetic field.

Substitute the known values and solve for mass

Plug in the known values into the equation: \((11.17C)(3131\frac{m}{s})(0.8T) = \frac{m(3131\frac{m}{s})^2}{2.015m}\) Solve for mass, m: \(m = \frac{(11.17C)(3131\frac{m}{s})(0.8T)(2.015m)}{(3131\frac{m}{s})^2}\) Calculate the mass: \(m \approx 0.03612 \, kg\) The mass of the aluminum ball is approximately \(0.03612 \, kg\).