Magnetic force on a current-carrying conductor in a uniform field (direction and magnitude) A straight conductor carries 10 A along the +x direction within a uniform magnetic flux density of 0.2 T directed along +z. For 1 m of conductor, find the magnitude and direction of the magnetic force.

Introduction / Context:The Lorentz force on a current-carrying segment in a magnetic field is foundational in electromagnetics and electric machines. This problem reinforces right-hand rule vector products and the linear dependence of force on current, field, and length.

Given Data / Assumptions:

• Current I = 10 A along +x.
• Magnetic flux density B = 0.2 T along +z.
• Conductor length considered L = 1 m oriented along +x.

Concept / Approach:Use the vector form of magnetic force on a current element: F = I * (L × B). With L along +x and B along +z, evaluate the cross product. Magnitude is |F| = I * L * B * sin(90°) because L is perpendicular to B. Direction follows the right-hand rule for L × B.

Step-by-Step Solution:Compute magnitude: |F| = 10 * 1 * 0.2 = 2 N.Determine direction: x × z = −y (since z × x = +y, then x × z = −y).Therefore, F = 2 N in the −y direction.

Verification / Alternative check:Apply Fleming’s left-hand rule: First finger (B) along +z, second finger (I) along +x, thumb points to −y, confirming the direction.

Why Other Options Are Wrong:“+y” reverses the cross-product orientation. Options with 0.02 N miscalculate magnitude by a factor of 100. “Zero force” is incorrect because current is perpendicular to the field, which maximizes force, not minimizes it.

Common Pitfalls:

• Swapping the order in the cross product (L × B ≠ B × L).
• Using degrees instead of radians improperly in vector formulas; here the sine factor is 1.

Final Answer:2 N, −y direction