Math  /  Algebra

Question7. Calculate the radius of the circular path taken by an alpha particle of charge 3.2×1019C3.2 \times 10^{-19} \mathrm{C} and a mass of 6.7×1027 kg6.7 \times 10^{-27} \mathrm{~kg} injected at a speed of 1.5×107 m/s1.5 \times 10^{7} \mathrm{~m} / \mathrm{s} into a uniform magnetic field of 2.4 T , at right angles to the field. (4 marks)

Studdy Solution

STEP 1

What is this asking? We need to find the radius of the circle an alpha particle makes when it enters a magnetic field.
It's like figuring out how tight a curve a car takes when entering a slippery patch of ice! Watch out! Make sure to use the correct values for the charge and mass of the alpha particle, and remember they're entering the field at a right angle!
This is crucial for our calculations.

STEP 2

1. Force Balance
2. Solve for Radius

STEP 3

Alright, let's **start** with the forces acting on this little alpha particle.
We know that the magnetic force is making it move in a circle, right?
That force, the **magnetic force**, is given by FB=qvBF_B = qvB, where qq is the **charge**, vv is the **velocity**, and BB is the **magnetic field strength**.

STEP 4

Now, since it's moving in a circle, there's also a **centripetal force** at play, pulling it inwards.
This force is given by Fc=mv2rF_c = \frac{mv^2}{r}, where mm is the **mass** and rr is the **radius** of the circle, which is what we're looking for!

STEP 5

Since these two forces are balancing each other out, we can set them equal to each other: qvB=mv2rqvB = \frac{mv^2}{r}.
This is a **key step**, as it connects the magnetic force to the circular motion.

STEP 6

Now, we want to find the **radius**, rr, so let's **rearrange** our equation.
We can multiply both sides by rr and divide by qvBqvB to get r=mv2qvBr = \frac{mv^2}{qvB}.
Notice how neatly the vv terms simplify!
One vv on top and one on the bottom divide to one, leaving us with r=mvqBr = \frac{mv}{qB}.

STEP 7

Time to **plug in** our values!
We have m=6.7×1027kgm = \mathbf{6.7 \times 10^{-27} \, kg}, v=1.5×107m/sv = \mathbf{1.5 \times 10^7 \, m/s}, q=3.2×1019Cq = \mathbf{3.2 \times 10^{-19} \, C}, and B=2.4TB = \mathbf{2.4 \, T}.
Substituting these values into our equation gives us: r=(6.7×1027kg)(1.5×107m/s)(3.2×1019C)(2.4T)r = \frac{(6.7 \times 10^{-27} \, kg) \cdot (1.5 \times 10^7 \, m/s)}{(3.2 \times 10^{-19} \, C) \cdot (2.4 \, T)}

STEP 8

Let's **crunch those numbers**! r=10.05×1020kgm/s7.68×1019CTr = \frac{10.05 \times 10^{-20} \, kg \cdot m/s}{7.68 \times 10^{-19} \, C \cdot T} r=1.30859×101mr = 1.30859 \times 10^{-1} \, mr0.13mr \approx \mathbf{0.13 \, m}

STEP 9

The radius of the circular path is approximately 0.13m\mathbf{0.13 \, m}.

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