4. A proton having a speed of 3.0 x 106 m/s in a direction perpendicular to a uniform magnetic field moves in a circle of radius 0.20 m within the field. What is the magnitude of the magnetic field? (The charge and mass of a proton are q = 1.60 x 10-19 C and m = 1.67 x 10-27 kg). a) 0.080 T b) 0.16 T c) 0.24 T d) 0.32 T e) 0.36 T

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**Problem:**

A proton having a speed of \(3.0 \times 10^6 \, \text{m/s}\) in a direction perpendicular to a uniform magnetic field moves in a circle of radius \(0.20 \, \text{m}\) within the field. What is the magnitude of the magnetic field?

(The charge and mass of a proton are \(q = 1.60 \times 10^{-19} \, \text{C}\) and \(m = 1.67 \times 10^{-27} \, \text{kg}\).)

**Options:**
a) \(0.080 \, \text{T}\)  
b) \(0.16 \, \text{T}\)  
c) \(0.24 \, \text{T}\)  
d) \(0.32 \, \text{T}\)  
e) \(0.36 \, \text{T}\)  

**Explanation:**

To solve this problem, we use the formula for the radius of the circular path of a charged particle in a magnetic field:

\[
r = \frac{mv}{qB}
\]

Where:
- \(r\) is the radius of the circle,
- \(m\) is the mass of the proton,
- \(v\) is the speed of the proton,
- \(q\) is the charge of the proton,
- \(B\) is the magnetic field.

Rearranging the formula to solve for \(B\):

\[
B = \frac{mv}{qr}
\]

Substitute the known values:

\[
B = \frac{(1.67 \times 10^{-27} \, \text{kg})(3.0 \times 10^6 \, \text{m/s})}{(1.60 \times 10^{-19} \, \text{C})(0.20 \, \text{m})}
\]

Calculate \(B\) to find the magnitude of the magnetic field.
Transcribed Image Text:**Problem:** A proton having a speed of \(3.0 \times 10^6 \, \text{m/s}\) in a direction perpendicular to a uniform magnetic field moves in a circle of radius \(0.20 \, \text{m}\) within the field. What is the magnitude of the magnetic field? (The charge and mass of a proton are \(q = 1.60 \times 10^{-19} \, \text{C}\) and \(m = 1.67 \times 10^{-27} \, \text{kg}\).) **Options:** a) \(0.080 \, \text{T}\) b) \(0.16 \, \text{T}\) c) \(0.24 \, \text{T}\) d) \(0.32 \, \text{T}\) e) \(0.36 \, \text{T}\) **Explanation:** To solve this problem, we use the formula for the radius of the circular path of a charged particle in a magnetic field: \[ r = \frac{mv}{qB} \] Where: - \(r\) is the radius of the circle, - \(m\) is the mass of the proton, - \(v\) is the speed of the proton, - \(q\) is the charge of the proton, - \(B\) is the magnetic field. Rearranging the formula to solve for \(B\): \[ B = \frac{mv}{qr} \] Substitute the known values: \[ B = \frac{(1.67 \times 10^{-27} \, \text{kg})(3.0 \times 10^6 \, \text{m/s})}{(1.60 \times 10^{-19} \, \text{C})(0.20 \, \text{m})} \] Calculate \(B\) to find the magnitude of the magnetic field.
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