Question

In a 1.05-T magnetic field directed vertically upward, a particle having a positive charge of magnitude 6.50 * 10^-6C and initially moving into the page at 3.75 m/s is deflected. Determine the magnitude and direction of the magnetic force on the particle.

Answer 1

Answer:

a. 25.59 μN b. The magnetic force is directed to the left.

Explanation:

a. The magnitude of the magnetic force

The magnetic force, F on the charge is

F = BqvsinФ where B = magnetic field strength = 1.05 T, q = charge = 6.50 × 10⁻⁶ C and v = speed of charge = 3.75 m/s and Ф = angle between B and v = 90° (since the charge is moving into the page and the magnetic field is directed vertically upwards and they are thus perpendicular to each other)  

So, substituting the values of the variables into the equation, we have

F = BqvsinФ

F = 1.05 T × 6.50 × 10⁻⁶ C × 3.75 m/s × sin90°

F = 25.59 × 10⁻⁶ N

F = 25.59 μN

b. The direction of the magnetic force on the particle

Using Fleming's right-hand rule, the first finger represents the magnetic field and the second finger the direction of the charge(or its velocity), then the thumb represents the direction of the magnetic force on the charge. Since the magnetic field is directed vertically upwards and the particle is directed into the page, the magnetic force which is represented by the thumb is directed to the left.

So, the magnetic force is directed to the left.