A proton having a speed of $3.0 \times 10 ^ { 6 } \mathrm {~m} / \mathrm { s }$ in a direction perpendicular to a uniform magnetic field moves in a circle of radius $0.20 \mathrm {~m}$ within the field. What is the magnitude of the magnetic field? $( e$ $= 1.60 \times 10 ^ { - 19 } \mathrm { C } , m _ { \text {proton } } = 1.67 \times 10 ^ { - 27 } \mathrm {~kg}$ )
A) $0.16 \mathrm {~T}$
B) $0.080 \mathrm {~T}$
C) $0.24 \mathrm {~T}$
D) $0.36 \mathrm {~T}$
E) $0.32 \mathrm {~T}$

Question

Quizplus · Accepted Answer

The magnitude of the magnetic field can be found using the formula for the magnetic force acting on a moving charge, which equals the centripetal force required to keep the proton moving in a circle: $qvB = \frac{mv^2}{r}$. Solving for $B$, we get $B = \frac{mv}{qr}$. Substituting the given values: $B = \frac{(1.67 \times 10^{-27} \, \text{kg})(3.0 \times 10^6 \, \text{m/s})}{(1.60 \times 10^{-19} \, \text{C})(0.20 \, \text{m})} = 0.16 \, \text{T}$.

A Proton Having a Speed Of 3.0×106 m/s3.0 \times 10 ^ { 6 } \mathrm {~m} / \mathrm { s }3.0×106 m/s

A Proton Having a Speed Of $3.0 \times 10 ^ { 6 } \mathrm {~m} / \mathrm { s }$