Askiitians Tutor Team
Last Activity: 15 Days ago
To tackle the problem of the deflection of a charged bullet in a magnetic field, we first need to understand the forces acting on the bullet due to its charge and the magnetic field. Then, we can move on to the second part regarding the tension in the threads of the suspended rod PQ when a current flows through it.
Deflection of the Charged Bullet
When a charged particle moves through a magnetic field, it experiences a magnetic force that can cause it to deflect. The formula for the magnetic force (F) acting on a charged particle is given by:
F = q * v * B * sin(θ)
Where:
- q = charge of the particle (in coulombs)
- v = velocity of the particle (in meters per second)
- B = magnetic field strength (in teslas)
- θ = angle between the velocity vector and the magnetic field (in degrees)
In this case, the bullet has a charge of 4 C, a speed of 270 m/s, and is moving horizontally while the magnetic field is vertical (500 T). Since the bullet is moving perpendicular to the magnetic field, θ = 90 degrees, and sin(90) = 1. Thus, the magnetic force simplifies to:
F = q * v * B
Substituting the values:
F = 4 C * 270 m/s * 500 T
F = 540,000 N
This force will cause the bullet to deflect. To find the deflection (d) as it travels through a distance (s) of 100 m, we can use the formula for the deflection in a uniform magnetic field:
d = (F * s) / (m * g)
However, we need the mass of the bullet to proceed. Assuming the bullet has a mass (m) of 0.01 kg (10 grams), and using the acceleration due to gravity (g = 9.81 m/s²), we can calculate:
d = (540,000 N * 100 m) / (0.01 kg * 9.81 m/s²)
d = 550,000,000 m
This value seems excessively large, indicating that the mass of the bullet is likely much larger than assumed. The actual deflection would depend on the bullet's mass, which should be provided for a more accurate calculation.
Tension in the Suspended Rod PQ
Now, let’s analyze the tension in the threads supporting the rod PQ. When the switch S is open, there is no current flowing through the rod, and the only forces acting on it are its weight and the tension in the threads. The weight (W) of the rod can be calculated as:
W = m * g
Given the mass of the rod is 200 g (0.2 kg), we find:
W = 0.2 kg * 9.81 m/s² = 1.962 N
Since the rod is in equilibrium when the switch is open, the tension (T) in the threads must equal the weight:
T = W = 1.962 N
When the switch is closed, a current of 2.0 A flows through the rod PQ in a magnetic field of 0.5 T. The magnetic force (F_m) acting on the rod can be calculated using:
F_m = I * L * B
Where:
- I = current (in amperes)
- L = length of the rod (in meters)
- B = magnetic field strength (in teslas)
Substituting the values:
F_m = 2.0 A * 0.2 m * 0.5 T = 0.2 N
Now, when the switch is closed, the total force acting on the rod is the sum of the weight and the magnetic force. The tension in the threads must counteract both forces:
T = W + F_m
T = 1.962 N + 0.2 N = 2.162 N
Thus, when the switch is closed, the tension in the threads becomes 2.162 N. This analysis shows how magnetic forces can influence the equilibrium of a system, demonstrating the interplay between electricity and magnetism.