A small aluminum ball with a mass of \(5.00 \mathrm{~g}\) and a charge of \(15.0 \mathrm{C}\) is moving northward at \(3000 \mathrm{~m} / \mathrm{s}\). You want the ball to travel in a horizontal circle with a radius of \(2.00 \mathrm{~m}\), in a clockwise sense when viewed from above. Ignoring gravity, what is the magnitude and the direction of the magnetic field that must be applied to the aluminum ball to cause it to have this motion?

Short Answer

Expert verified
Answer: The magnitude of the applied magnetic field is 0.50 T, and the direction is downward (southward).

Step by step solution

01

Analyze the conditions and forces involved

The aluminum ball needs to travel in a horizontal circular path. The only force acting on it is the magnetic force, which also provides the centripetal force for circular motion.
02

Write down the magnetic force equation and centripetal force equation

The magnetic force acting on a charged particle can be described by the Lorentz force equation: $$F_B = qvB\sin(\theta)$$ where \(F_B\) is the magnetic force, \(q\) is the charge of the particle, \(v\) is its velocity, \(B\) is the magnetic field, and \(\theta\) is the angle between the velocity and magnetic field vectors. The centripetal force needed for circular motion is given by: $$F_c = \frac{mv^2}{r}$$ where \(F_c\) is the centripetal force, \(m\) is the mass of the particle, \(v\) is its velocity, and \(r\) is the radius of the circular path.
03

Equate magnetic and centripetal forces

Since the magnetic force provides the centripetal force for circular motion, we can set the two forces equal to each other: $$qvB\sin(\theta) = \frac{mv^2}{r}$$
04

Solve for the magnitude of the magnetic field

We are asked to find the magnitude \(B\) of the applied magnetic field. Rearrange the equation from the previous step to solve for \(B\): $$B = \frac{mv^2}{rqv\sin(\theta)}$$ Plug in the given values for mass (\(m = 5.00 \mathrm{~g} = 0.005 \mathrm{~kg}\)), charge (\(q = 15.0 \mathrm{C}\)), velocity (\(v = 3000 \mathrm{~m} / \mathrm{s}\)), and radius (\(r = 2.00 \mathrm{~m}\)). Since the ball is supposed to travel clockwise when viewed from above, the magnetic field should be pointing downward (southward), and the angle between the magnetic field and the velocity vector is \(\theta = 90^\circ\). Therefore, \(\sin(\theta) = 1\). $$B = \frac{(0.005 \mathrm{~kg})(3000 \mathrm{~m} / \mathrm{s})^2}{(2.00 \mathrm{~m})(15.0 \mathrm{C})(3000 \mathrm{~m} / \mathrm{s})(1)}$$ $$B = \frac{(0.005)(9\cdot 10^6)}{(2)(15)(3\cdot 10^3)}$$ $$B = \frac{45\cdot 10^3}{90\cdot 10^3}$$ $$B = 0.50 \mathrm{T}$$
05

Determine the direction of the magnetic field

As mentioned earlier, for the aluminum ball to travel clockwise when viewed from above, the magnetic field should be pointing downward (southward).
06

Write the final answer

Therefore, the magnitude and direction of the magnetic field that must be applied to the aluminum ball for the desired motion are \(0.50 \mathrm{T}\) and downward (southward), respectively.

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