Test on Magnetic field of a current-carrying wire

Question 1: Two current-carrying wires placed parallel to each other will exert a magnetic force on one another.

A. Yes

B. No

Explanation: Each current-carrying wire produces a magnetic field. When another current-carrying wire is placed in this magnetic field, it experiences a force. Therefore, two parallel current-carrying wires will exert a magnetic force on each other due to the interaction of their respective magnetic fields and currents.

Question 2: The magnetic field strength (B) at a point near a long, straight current-carrying wire is inversely proportional to the perpendicular distance (r) from the wire to that point.

A. Yes

B. No

Explanation: The study materials state that the formula for the magnetic field strength B is μ₀·I / (2π·r). This formula explicitly shows that B is inversely proportional to r, the distance from the wire. As r increases, B decreases proportionally.

Question 3: A long, straight wire carries a steady electric current of 8.0 A. What is the magnitude of the magnetic field strength at a point 20.0 cm away from the wire?

A. 4.0 × 10⁻⁶ T

B. 8.0 × 10⁻⁶ T

C. 1.6 × 10⁻⁵ T

D. 8.0 × 10⁻⁷ T

Explanation: The magnetic field strength (B) at a distance (r) from a long, straight current-carrying wire is given by the formula B = μ₀·I / (2π·r). Given a current (I) of 8.0 A and a distance (r) of 20.0 cm (which is 0.20 m), and knowing that the permeability of free space (μ₀) is 4π·10⁻⁷ T·m/A, we can substitute these values: B = (4π·10⁻⁷ T·m/A · 8.0 A) / (2π · 0.20 m) B = (32π·10⁻⁷ T·m) / (0.40π m) B = (32 / 0.40) · 10⁻⁷ T B = 80 · 10⁻⁷ T B = 8.0 · 10⁻⁶ T. Therefore, the magnetic field strength is 8.0 × 10⁻⁶ Tesla.

Question 4: Two long, parallel wires are separated by a distance of 5.0 cm. Wire 1 carries a current of 3.0 A, and Wire 2 carries a current of 7.0 A. If the currents flow in the same direction, what is the force per unit length between the wires?

A. 8.4 · 10⁻⁵ N/m

B. 8.4 · 10⁻⁷ N/m

C. 4.2 · 10⁻⁵ N/m

D. 1.2 · 10⁻⁴ N/m

Explanation: The force per unit length (F/L) between two parallel current-carrying wires is given by the formula F/L = (μ₀·I₁·I₂) / (2π·d). Given μ₀ = 4π·10⁻⁷ T·m/A, I₁ = 3.0 A, I₂ = 7.0 A, and d = 5.0 cm = 0.05 m. Substituting these values: F/L = (4π·10⁻⁷ T·m/A · 3.0 A · 7.0 A) / (2π · 0.05 m) F/L = (84π·10⁻⁷ T·m·A) / (0.10π m) F/L = (84 / 0.10) · 10⁻⁷ N/m F/L = 840 · 10⁻⁷ N/m F/L = 8.4 · 10⁻⁵ N/m. Since the currents are in the same direction, the force is attractive.

Question 5: A magnetic field simulation helps students gain an intuitive understanding of electromagnetic waves and their propagation.

A. Yes

B. No

Explanation: The study materials state that by manipulating parameters in a magnetic field simulation, 'you gain an intuitive understanding of these fundamental principles and their practical implications,' referring to magnetic fields, forces, Ampere's Law, the right-hand rule, field strength, and superposition. The provided text does not mention electromagnetic waves or their propagation as a concept for which the simulation provides intuitive understanding.