Questions: Force on Current-Carrying Conductors in Magnetic Fields

Question 1

A straight wire carrying current I is oriented parallel to a uniform magnetic field B. What is the magnetic force on the wire?

• A. F = ILB, directed perpendicular to both the wire and the field
• B. F = ILB, directed along the wire
• C. F = 0, because the cross product of parallel vectors is zero
• D. F = ILB/2, because parallel orientation gives half the maximum force

Answer: (C)

The force is F = IL × B. The magnitude is ILB sin θ, where θ is the angle between the current direction and B. When the wire is parallel to B, θ = 0° (or 180°), so sin θ = 0 and F = 0. Physically, the conduction electrons drift along the field direction, so v ∥ B, and the cross product v × B = 0 — the Lorentz force on each charge is zero. Maximum force occurs at θ = 90° (wire perpendicular to field).

Question 2

A rectangular current loop lies entirely within a uniform magnetic field. What can you conclude about the net force on the loop?

• A. The net force is nonzero and directed toward the stronger part of the field
• B. The net force is zero, because opposite sides of the loop carry currents in opposite directions and their forces cancel
• C. The net force is zero, which follows from ∮ dl = 0 for any closed path
• D. The net force equals ILB for the longest side of the loop

Answer: (C)

Both B and C are essentially correct statements (opposite currents + the mathematical identity), but C captures the deeper reason more precisely. For any closed loop in a uniform field: F = ∮ I(dl × B) = I(∮ dl) × B = I(0) × B = 0, because ∮ dl = 0 (the loop returns to its starting point). This result is independent of the loop's shape. Importantly, the net force is zero but the loop may still experience a nonzero *torque*, which is the basis of electric motor operation.

Question 3

The force on a current-carrying wire is greatest when the wire is oriented perpendicular to the magnetic field.

• T. True
• F. False

Answer: True

The magnitude of the force is F = ILB sin θ, where θ is the angle between the current direction and the field B. sin θ is maximized at θ = 90° (perpendicular orientation), giving F_max = ILB. At θ = 0° (parallel), sin θ = 0 and F = 0. This follows directly from the cross product: |L × B| = LB sin θ, which peaks when L and B are perpendicular.

Question 4

A closed current loop in a perfectly uniform magnetic field experiences both a net force and a net torque.

• T. True
• F. False

Answer: False

In a uniform field, the net *force* on a closed loop is always zero (because ∮ dl = 0 for any closed path, so F = I(∮ dl) × B = 0). However, the loop generally experiences a nonzero *torque* (unless the loop's magnetic moment is aligned with the field). Torque and force are independent: zero net force does not imply zero torque. This distinction is critical for understanding electric motors, which rotate due to torque despite experiencing no net translational force in a uniform field.

Question 5

Starting from the Lorentz force law on a single charge, derive the force formula F = IL × B for a straight current-carrying wire segment of length L.

Model answer: Consider a wire segment of length L, cross-sectional area A, with n charge carriers per unit volume, each with charge q and drift velocity v_d. The total number of charges in the segment is nAL. The Lorentz force on each charge is F_single = qv_d × B. The total force on the segment is (nAL)(qv_d × B) = (nqv_dA)L × B. But nqv_dA is exactly the definition of current I (charge per unit time crossing a cross-section). Substituting: F = IL × B, where L is a vector of magnitude L pointing in the direction of conventional current flow.

The key step is recognizing that current I = nqv_dA consolidates all the microscopic details (carrier density, charge, drift speed, area) into one macroscopic quantity. Once you make this substitution, the formula F = IL × B is nothing more than the Lorentz force on all carriers rewritten in terms of the measurable current. This derivation also shows why the force is zero when the wire is parallel to B: in that case, v_d ∥ B, so v_d × B = 0 for each individual carrier.