Force on a Wire in a Vector Magnetic Field B = 2i + 3j - 4k
Physics · Moving Charges And Magnetism · NEET
The force on a straight current wire in a magnetic field is F = I (L x B), where L is the length vector pointing along the current direction. When B is given in i, j, k form, you take the cross product of L with B, then find the magnitude. Memory hook: "Point L along the current, cross it with B, take the root of the squares." For a wire of length L along +x in B = 2i + 3j - 4k, only the j and k parts of B matter, giving |F| = 5 I L.
The wire lies along +x, so the parallel 2i part of B gives zero force; the perpendicular 3j and -4k parts combine to give |F| = 5 I L.
Your doubts, answered
If the wire is along the x-axis, why does the 2i part of B not matter?
Force needs L x B. The wire vector is L = L i. The cross product i x i = 0, so the 2i part of B (which is parallel to the wire) contributes nothing. Only the parts of B perpendicular to the wire (3j and -4k) create force. Rule: the component of B along the wire does no work of pushing it sideways.
How do I take the magnitude after the cross product?
After L x B you get a vector like (4j + 3k) times I L. The magnitude is I L times the square root of the sum of squares of the components. Here sqrt(4^2 + 3^2) = sqrt(25) = 5, so |F| = 5 I L. Never add components directly; always use the root of the squares.
What are the unit vector cross product rules I keep forgetting?
Follow the cyclic order i -> j -> k -> i. Going forward is positive: i x j = k, j x k = i, k x i = j. Going backward is negative: j x i = -k, i x k = -j, k x j = -i. Any vector crossed with itself is zero: i x i = j x j = k x k = 0.
Is L a vector or just a number in F = I L x B?
L is a vector. Its magnitude is the wire length and its direction is the direction of conventional current in the wire. That is why a wire carrying current along +x has L = L i, not just L. Getting this direction wrong flips or zeros your answer.
⚠️ The NEET trap ✗ Using all three components of B, so |F| = I L sqrt(2^2 + 3^2 + 4^2) = I L sqrt(29). ✓ Only the perpendicular parts survive the cross product. With L along +x, the 2i term drops out, giving |F| = I L sqrt(4^2 + 3^2) = 5 I L. 🧠 The part of B pointing the same way as the wire does nothing. Cross out the parallel component before you square anything.
Real NEET questions
2023
A wire carrying a current I along the positive x-axis has length L. It is kept in a magnetic field B = (2i + 3j - 4k) T. The magnitude of the magnetic force acting on the wire is:
A · 3IL
B · 5 sqrt(IL)
C · 5IL ✓
D · sqrt(3) IL
Solution: The wire vector points along the current: L = L i. Force F = I (L x B) = I L i x (2i + 3j - 4k). Use i x i = 0, i x j = k, i x k = -j. So F = I L [2(0) + 3(k) - 4(-j)] = I L (4j + 3k). Magnitude |F| = I L sqrt(4^2 + 3^2) = I L sqrt(25) = 5 I L. Correct option: C.
Solved Moving Charges And Magnetism NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What is the formula for force on a current-carrying wire?
F = I (L x B). I is the current, L is the length vector along the current direction, and B is the magnetic field. The magnitude in the simple case is F = B I L sin(theta), where theta is the angle between the wire and B.
Why is the answer 5IL and not sqrt(29) I L for B = 2i + 3j - 4k?
Because the wire is along +x. The 2i part of B is parallel to the wire and gives zero force (i x i = 0). Only 3j and -4k contribute, so |F| = I L sqrt(3^2 + 4^2) = 5 I L.
How do I decide the direction of the force vector?
After computing L x B you get a vector, for example I L (4j + 3k). That vector direction is the force direction. You do not need the right-hand rule separately once you have done the full cross product.
Does the length of the wire change the direction of the force?
No. Length only scales the magnitude. The direction of the force depends only on the current direction (L direction) and the field B through the cross product.
