Q30. An alpha-particle moves with a speed of (5 x 10⁵î)ms⁻¹. It enters a region where there is a mag

Q: What is the correct answer for CUET UG Physics Q30?

The correct answer is c. The magnetic force (Lorentz force) on a charged particle moving in a magnetic field is given by F = qvBsinθ, where q is the charge of the particle, v is its speed, B is the magnetic field strength, and θ is the angle between the velocity vector (v) and the magnetic field vector (B). For an alpha particle, the charge q = +2e = 2 × (1.6 × 10⁻¹⁹ C) = 3.2 × 10⁻¹⁹ C. The speed v = 5 × 10⁵ m/s (along th

Q: Q30. An alpha-particle moves with a speed of (5 x 10⁵î)ms⁻¹. It enters a region where there is a magnetic field of magnitude 4T, directed at an angle of 45° to the X-axis and lying in XY plane. The…

The correct answer is c. The magnetic force (Lorentz force) on a charged particle moving in a magnetic field is given by F = qvBsinθ, where q is the charge of the particle, v is its speed, B is the magnetic field strength, and θ is the angle between the velocity vector (v) and the magnetic field vector (B). For an alpha particle, the charge q = +2e = 2 × (1.6 × 10⁻¹⁹ C) = 3.2 × 10⁻¹⁹ C. The speed v = 5 × 10⁵ m/s (along th

Q: What are the options for this CUET UG Physics question?

A. 1.6 x 10⁻¹⁴ N · B. 3.2 x 10⁻¹³ N · C. 4.5 x 10⁻¹³ N · D. 1.6 x 10⁻¹³ N

Question

Q30. An alpha-particle moves with a speed of (5 x 10⁵î)ms⁻¹. It enters a region where there is a magnetic field of magnitude 4T, directed at an angle of 45° to the X-axis and lying in XY plane. The magnitude of magnetic force on alpha particle is:
(a) 1.6 x 10⁻¹⁴ N
(b) 3.2 x 10⁻¹³ N
(c) 4.5 x 10⁻¹³ N
(d) 1.6 x 10⁻¹³ N

FirstInTest · Accepted Answer

The magnetic force (Lorentz force) on a charged particle moving in a magnetic field is given by F = qvBsinθ, where q is the charge of the particle, v is its speed, B is the magnetic field strength, and θ is the angle between the velocity vector (v) and the magnetic field vector (B). For an alpha particle, the charge q = +2e = 2 × (1.6 × 10⁻¹⁹ C) = 3.2 × 10⁻¹⁹ C. The speed v = 5 × 10⁵ m/s (along the x-axis). The magnetic field B = 4 T, and the angle θ between the velocity (x-axis) and the magnetic field is 45°. Plugging these values into the formula: F = (3.2 × 10⁻¹⁹ C) × (5 × 10⁵ m/s) × (4 T) × sin(45°). F = (3.2 × 5 × 4) × 10⁻¹⁴ × (1/√2) N = 64 × 10⁻¹⁴ / √2 N = 32√2 × 10⁻¹⁴ N. Since √2 ≈ 1.414, F ≈ 32 × 1.414 × 10⁻¹⁴ N = 45.248 × 10⁻¹⁴ N = 4.5248 × 10⁻¹³ N. This value is approximately 4.5 × 10⁻¹³ N, matching option (c).

Answer

1.6 x 10⁻¹⁴ N

Answer

3.2 x 10⁻¹³ N

Answer

4.5 x 10⁻¹³ N

Answer

1.6 x 10⁻¹³ N

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