A charged particle moves at 2.5 × 10⁴ m/s at an angle of 25° to a magnetic field that has a field strength of 8.1 × 10⁻² T. If the magnetic force is 7.5 × 10⁻² N, what is the magnitude of the charge?
Options:
A) 3.7 × 10⁻⁵ C
B) 4.1 × 10⁻⁵ C
C) 8.8 × 10⁻⁵ C
D) 1.0 × 10⁻⁴ C
Answer (2)
To find the magnitude of the charge of the particle, we can use the formula for the magnetic force on a moving charge:
F = q v B sin ( θ )
where:
F is the magnetic force (7.5 \times 10^{-2} \text{ N})
q is the magnitude of the charge (which we want to find)
v is the velocity of the charged particle (2.5 \times 10^4 \text{ m/s})
B is the magnetic field strength (8.1 \times 10^{-2} \text{ T})
θ is the angle between the velocity and the magnetic field (25°)
We can rearrange the formula to solve for q :
q = v B sin ( θ ) F
First, calculate sin ( 25° ) :
sin ( 25° ) ≈ 0.4226
Now plug the known values into the equation:
q = 2.5 × 1 0 4 × 8.1 × 1 0 − 2 × 0.4226 7.5 × 1 0 − 2
Calculate the denominator:
2.5 × 1 0 4 × 8.1 × 1 0 − 2 × 0.4226 ≈ 855.465
Now divide the values:
q = 855.465 7.5 × 1 0 − 2 ≈ 8.8 × 1 0 − 5 C
Thus, the magnitude of the charge is 8.8 × 1 0 − 5 C .
The correct answer is Option C: 8.8 \times 10^{-5} \text{ C}.
The magnitude of the charge of the particle is approximately 8.8 × 10⁻⁵ C, calculated using the formula for magnetic force on a moving charge. Thus, the correct answer is Option C: 8.8 × 10⁻⁵ C.
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