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The path of an electron, traveling at [tex]$3.7 \times 10^4$[/tex] meters/second, is making a right angle to a magnetic field. A single electron experiences a force of [tex]$2.2 \times 10^{-15}$[/tex] newtons while in the field. What is the strength of the magnetic field? The value of [tex][tex]$q = -1.6 \times 10^{-19}$[/tex][/tex] coulombs.

A. 0.16 teslas
B. 0.22 teslas
C. 0.37 teslas
D. 0.73 teslas


Sagot :

To determine the strength of the magnetic field, we use the formula for the magnetic force experienced by a charged particle moving perpendicular to the magnetic field:

[tex]\[ F = q \cdot v \cdot B \][/tex]

Where:
- [tex]\( F \)[/tex] is the magnetic force,
- [tex]\( q \)[/tex] is the charge of the particle,
- [tex]\( v \)[/tex] is the velocity of the particle,
- [tex]\( B \)[/tex] is the magnetic field strength (which we need to find).

Given values:
- Force ([tex]\( F \)[/tex]) = [tex]\( 2.2 \times 10^{-15} \)[/tex] newtons,
- Charge ([tex]\( q \)[/tex]) = [tex]\( -1.6 \times 10^{-19} \)[/tex] coulombs,
- Velocity ([tex]\( v \)[/tex]) = [tex]\( 3.7 \times 10^4 \)[/tex] meters/second.

First, rearrange the formula to solve for [tex]\( B \)[/tex]:

[tex]\[ B = \frac{F}{q \cdot v} \][/tex]

Now, substitute the given values into the formula:

[tex]\[ B = \frac{2.2 \times 10^{-15}}{-1.6 \times 10^{-19} \cdot 3.7 \times 10^4} \][/tex]

Perform the calculation inside the denominator first:

[tex]\[ q \cdot v = -1.6 \times 10^{-19} \cdot 3.7 \times 10^4 = -5.92 \times 10^{-15} \][/tex]

Next, divide the force by this result:

[tex]\[ B = \frac{2.2 \times 10^{-15}}{-5.92 \times 10^{-15}} = -0.3716216216216216 \][/tex]

Since the question asks for the magnitude of the magnetic field, we take the absolute value:

[tex]\[ |B| = 0.3716216216216216 \][/tex]

So, the strength of the magnetic field is approximately [tex]\( 0.37 \)[/tex] teslas. Therefore, the correct answer is:

C. 0.37 teslas