A point charge of 5.0 × 10⁻⁷ C moves to the right at 2.6 × 10⁵ m/s in a magnetic field that is directed into the screen and has a field strength of 1.8 × 10⁻² T. What is the magnitude of the magnetic force acting on the charge?
Options:
0 N
2.3 × 10⁻³ N
23 N
2.3 × 10¹¹ N
Answer (2)
To solve this problem, we use the formula for the magnetic force acting on a moving charge in a magnetic field:
F = q v B sin ( θ )
Where:
F is the magnetic force,
q is the charge, 5.0 × 1 0 − 7 C,
v is the velocity, 2.6 × 1 0 5 m/s,
B is the magnetic field strength, 1.8 × 1 0 − 2 T,
θ is the angle between the velocity of the charge and the magnetic field direction.
In this problem, the magnetic field is described as being directed 'into the screen', and the charge is moving 'to the right'. This means that the angle θ between the velocity of the charge and the magnetic field is 90 degrees, or π /2 radians. The sine of 90 degrees is 1.
Therefore, the calculation simplifies to:
F = ( 5.0 × 1 0 − 7 C ) ( 2.6 × 1 0 5 m/s ) ( 1.8 × 1 0 − 2 T ) ( 1 )
Calculating:
F = 5.0 × 2.6 × 1.8 × 1 0 − 7 + 5 − 2 F = 23.4 × 1 0 − 4 F = 2.34 × 1 0 − 3 N
Rounding to two significant figures, the magnitude of the magnetic force is 2.3 × 1 0 − 3 N .
Therefore, the correct option is:
2.3 \times 10^{-3} \text{ N}
The magnetic force acting on the charge can be calculated using the formula F = q v B sin ( θ ) . After calculating with the given values, the magnitude of the force is determined to be 2.3 × 1 0 − 3 N . Therefore, the correct option is 2.3 × 10⁻³ N .
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