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
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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Question
![**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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd9a2776d-a528-4478-be0e-235d689690c2%2F12d5c83a-67f0-47dd-89bf-8dd97ec1d61f%2F3cy8tw_processed.jpeg&w=3840&q=75)
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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