Question

A particle has a charge of 5.2 × 10-19 coulombs and experiences a force of 9.5 × 10-15 newtons when it travels through a magnetic field with strength 2.2 × 10-1 tesla. What is the speed of the particle?

A.
1.9 × 104 meters/second
B.
2.2 × 104 meters/second
C.
4.9 × 104 meters/second
D.
8.3 × 104 meters/second

Answer 1

To find the speed of the particle when it is in a magnetic field, you can use the formula for the force experienced by a charged particle moving in a magnetic field:

\[ F = qvB \]

Here:
- \( F \) is the force on the particle (in newtons),
- \( q \) is the charge of the particle (in coulombs),
- \( v \) is the speed of the particle (in meters/second),
- \( B \) is the magnetic field strength (in tesla).

We can rearrange this formula to solve for the speed \( v \):

\[ v = \frac{F}{qB} \]

Given:
- \( F = 9.5 \times 10^{-15} \, \text{N} \)
- \( q = 5.2 \times 10^{-19} \, \text{C} \)
- \( B = 2.2 \times 10^{-1} \, \text{T} \)

Now substituting the values into the equation:

\[
v = \frac{9.5 \times 10^{-15}}{(5.2 \times 10^{-19})(2.2 \times 10^{-1})}
\]

Calculating the denominator:

\[
(5.2 \times 10^{-19}) \times (2.2 \times 10^{-1}) = 1.144 \times 10^{-19} \, \text{C} \cdot \text{T}
\]

Now substitute this back into the velocity equation:

\[
v = \frac{9.5 \times 10^{-15}}{1.144 \times 10^{-19}}
\]

Calculating \( v \):

\[
v \approx 8.30 \times 10^{4} \, \text{m/s}
\]

Thus, the speed of the particle is approximately \( 8.3 \times 10^{4} \, \text{m/s} \).

The correct answer is:

**D. 8.3 × 10^4 meters/second**