!!!!!!I am unable to crack yhis one. IT IS VERY TRICKY AND TACTICAL!

(1.40T)i and (b) B = (1.4OT)k? 27.2. A particle of mass 0.195 g carries a charge of -2.50 X 10^- 8 C. The particle is given an initial horizontal velocity that is due north and has magnitude 4.00 X 10^ 4 m/s. What are the magni- tude and direction of the rninirnum magnetic field that will keep the particle moving in the earth's gravitational field in the same horizontal, northward direction?

To solve this problem, we need to analyze the forces acting on the charged particle and how a magnetic field can influence its motion. The particle is moving in a gravitational field, and we want to find the magnetic field that will keep it moving horizontally in the same direction. Let's break this down step by step.

Understanding the Forces at Play

The particle has a mass of 0.195 g, which we need to convert to kilograms for our calculations:

• Mass (m) = 0.195 g = 0.195 / 1000 = 0.000195 kg

The charge of the particle is given as -2.50 x 10^-8 C. Since the charge is negative, it will experience a force in the opposite direction of the magnetic field when it moves through it.

Gravitational Force

The gravitational force acting on the particle can be calculated using the formula:

• F_gravity = m * g

Where g is the acceleration due to gravity (approximately 9.81 m/s²).

• F_gravity = 0.000195 kg * 9.81 m/s² ≈ 0.00191595 N

Magnetic Force

The magnetic force acting on a charged particle moving in a magnetic field is given by:

• F_magnetic = q * v * B * sin(θ)

Where:

• q = charge of the particle
• v = velocity of the particle
• B = magnetic field strength
• θ = angle between the velocity and the magnetic field direction

In this case, since the particle is moving horizontally due north and we want the magnetic force to act vertically upward to balance the gravitational force, we can set θ = 90 degrees. Thus, sin(90°) = 1, simplifying our equation to:

• F_magnetic = q * v * B

Setting Forces Equal

To keep the particle moving in the same horizontal direction, the magnetic force must equal the gravitational force:

• q * v * B = F_gravity

Substituting Known Values

Now we can substitute the known values into the equation:

• q = -2.50 x 10^-8 C
• v = 4.00 x 10⁴ m/s
• F_gravity ≈ 0.00191595 N

Plugging these into the equation gives:

• -2.50 x 10^-8 C * 4.00 x 10⁴ m/s * B = 0.00191595 N

Solving for B

Now we can solve for B:

• B = 0.00191595 N / (-2.50 x 10^-8 C * 4.00 x 10⁴ m/s)

Calculating this gives:

• B = 0.00191595 / (-1.00 x 10^-3)
• B = -1.91595 T

Direction of the Magnetic Field

The negative sign indicates that the direction of the magnetic field is opposite to the direction of the velocity of the particle. Since the particle is moving north, the magnetic field must point downward (toward the ground) to exert an upward force on the negatively charged particle.

Final Answer

Thus, the magnitude of the minimum magnetic field required to keep the particle moving in the same horizontal, northward direction is approximately 1.92 T, directed downward.