Question

A proton moving with a speed of \(4.00 \cdot 10^{5} \mathrm{~m} / \mathrm{s}\) in the positive \(y\) -direction enters a uniform magnetic field of \(0.400 \mathrm{~T}\) pointing in the positive \(x\) -direction. Calculate the magnitude of the force on the proton.

Short answer

Answer: The magnitude of the force on the proton is \(2.56 \times 10^{-14}\,\mathrm{N}\).

Step-by-step solution

Identify the given variables

We are given the following information: - Speed of the proton, \(v = 4.00 \times 10^5 \mathrm{~m/s}\) - Magnetic field strength, \(B = 0.400 \mathrm{~T}\) - Proton's charge, \(q = +1.60217646 \times 10^{-19} \,\mathrm{C}\) (positive because it's a proton) - Direction of proton's movement: positive \(y\)-direction - Direction of magnetic field: positive \(x\)-direction

Find the angle between the proton's velocity and the magnetic field

Since the proton is moving in the positive \(y\)-direction and the magnetic field is in the positive \(x\)-direction, the angle between the two is 90 degrees. Thus, \(\theta = 90^{\circ}\) or \(\theta = \frac{\pi}{2}\) radians.

Calculate the force on the proton using the formula

Now we will use the formula for magnetic force on a moving charge: \(F=qvB\sin\theta\) and plug in the given values. $$F = (1.60217646\times10^{-19}\,\mathrm{C})(4.00\times10^5\,\mathrm{m/s})(0.400\,\mathrm{T})\sin\left(\frac{\pi}{2}\right)$$ Since \(\sin(\frac{\pi}{2}) = 1\), we can simplify the expression for the force: $$F = (1.60217646\times10^{-19}\,\mathrm{C})(4.00\times10^5\,\mathrm{m/s})(0.400\,\mathrm{T})$$

Calculate the numerical value of the force

Perform the multiplication for the force: $$F = 2.562682336 \times 10^{-14}\,\mathrm{N}$$

Round the answer to an appropriate number of significant figures

Since both \(v\) and \(B\) have 3 significant figures, we will round the force value to 3 significant figures: $$F \approx 2.56 \times 10^{-14}\,\mathrm{N}$$ Therefore, the magnitude of the force on the proton is \(2.56 \times 10^{-14}\,\mathrm{N}\).