Magnetic Force on Wire - Physics 2

1. What is the magnitude of the magnetic force on this wire?
   Answer: $0.80 \text{ N}$, directed in the $+y$-direction, because only the $z$-component of $\vec{B}$ contributes to the cross product $\hat{x} \times \hat{z}$, yielding $F = ILB_z = (4.0)(0.50)(0.40)$. When a current-carrying wire sits in a magnetic field, the force on it comes from $\vec{F} = I\vec{L} \times \vec{B}$. The critical insight is that **only the component of $\vec{B}$ perpendicular to the current contributes** — a field component parallel to the current produces zero force, since $\hat{x} \times \hat{x} = 0$. Here, the current flows in the $+x$-direction, so $\vec{L} = L\hat{x} = 0.50\hat{x}$. Expanding the cross product: $\vec{F} = I\vec{L} \times \vec{B} = I(L\hat{x}) \times (0.30\hat{x} - 0.40\hat{z})$ $= IL\bigl[0.30(\hat{x}\times\hat{x}) - 0.40(\hat{x}\times\hat{z})\bigr]$ Since $\hat{x}\times\hat{x} = 0$ and $\hat{x}\times\hat{z} = -\hat{y}$: $\vec{F} = IL(0.40)\hat{y} = (4.0)(0.50)(0.40)\hat{y} = 0.80\hat{y} \text{ N}$ This confirms A is correct: the magnitude is $0.80 \text{ N}$ in the $+y$-direction, driven entirely by the $z$-component of $\vec{B}$. B is wrong because you cannot simply use $B_{\text{total}}$ in the force formula — the parallel component ($B_x$) is physically incapable of exerting a force on the wire, so lumping both components together overstates the result. C compounds that same error by adding a force from $B_x$, which produces no force at all. D introduces an unnecessary angle calculation and arrives at a value that matches neither the correct cross product nor the geometry. Your study tip: always decompose $\vec{B}$ and ask yourself which components are parallel vs. perpendicular to $\vec{L}$ before calculating. Parallel components vanish — only perpendicular components do work in the cross product.
2. What is the magnitude and direction of the magnetic force on this wire segment?
   Answer: $|\vec{F}| = ILB_0/\sqrt{2} \approx 0.85 \text{ N}$, directed in the $+\hat{z}$-direction, because $\vec{F} = IL(\hat{u}\times\vec{B}) = IL\left(\frac{1}{\sqrt{2}}\hat{x}+\frac{1}{\sqrt{2}}\hat{y}\right)\times(B_0\hat{y}) = \frac{ILB_0}{\sqrt{2}}(\hat{x}\times\hat{y}) = \frac{ILB_0}{\sqrt{2}}\hat{z}$, since $\hat{y}\times\hat{y}=0$. When a current-carrying wire sits in a magnetic field, the force on it is $\vec{F} = I\vec{L} \times \vec{B}$, where $\vec{L}$ is the length vector pointing in the direction of current flow. The key insight is that only the component of the wire perpendicular to $\vec{B}$ contributes to the force — parallel components produce zero force because $\hat{y} \times \hat{y} = 0$. Here, $\vec{L} = L\hat{u} = L\left(\frac{1}{\sqrt{2}}\hat{x} + \frac{1}{\sqrt{2}}\hat{y}\right)$ and $\vec{B} = B_0\hat{y}$. Computing the cross product: $\vec{F} = IL\left(\frac{1}{\sqrt{2}}\hat{x} + \frac{1}{\sqrt{2}}\hat{y}\right) \times B_0\hat{y} = \frac{ILB_0}{\sqrt{2}}(\hat{x}\times\hat{y}) + \frac{ILB_0}{\sqrt{2}}(\hat{y}\times\hat{y})$. Since $\hat{x}\times\hat{y} = +\hat{z}$ and $\hat{y}\times\hat{y} = 0$, the result is $\frac{ILB_0}{\sqrt{2}}\hat{z} \approx 0.85\ \text{N}$ in the $+\hat{z}$-direction. That confirms B. A is wrong not in magnitude but in direction — it claims $\hat{x}\times\hat{y} = -\hat{z}$, which is a right-hand rule error. Using the right-hand rule correctly, fingers curl from $\hat{x}$ toward $\hat{y}$ and the thumb points in $+\hat{z}$, not $-\hat{z}$. C ignores the wire's orientation entirely and incorrectly applies $F = ILB$ as if the wire were perpendicular to $\vec{B}$, overstating the force. D confuses which component matters — the parallel component (along $\hat{y}$) contributes nothing, and the direction of force must be perpendicular to both the wire and the field, never along $\hat{y}$. Your go-to strategy: always distribute the cross product term by term. Parallel terms vanish instantly, and perpendicular terms give you both the magnitude and direction cleanly — no guessing required.
3. Is the student's claim correct, and what is the magnetic force on the original wire?
   Answer: The claim is correct, and the original force is $F = ILB\sin30° = (8.0)(0.30)(0.25)(0.50) = 0.30 \text{ N}$, because $F \propto IL$ and doubling $I$ while halving $L$ keeps the product $IL$ constant at $2.4 \text{ A}\cdot\text{m}$. When a current-carrying wire sits in a magnetic field at an angle, the force on it follows $F = ILB\sin\theta$. The sine function appears because only the component of the wire perpendicular to the field contributes to the force — this is a direct consequence of the cross product in the Lorentz force law. For the original wire: $F = (8.0)(0.30)(0.25)\sin30° = (8.0)(0.30)(0.25)(0.50) = 0.30 \text{ N}$. Now consider the student's modification: double $I$ to $16.0 \text{ A}$, halve $L$ to $0.15 \text{ m}$. The new force is $F' = (16.0)(0.15)(0.25)\sin30° = (2.4)(0.25)(0.50) = 0.30 \text{ N}$. Since $IL$ stays at $2.4 \text{ A}\cdot\text{m}$ and $B$ and $\theta$ are unchanged, the force is indeed preserved. C is correct. Choice A contains a subtle trap — it claims that shortening the wire somehow changes the effective angle $\theta$. It doesn't. The angle between the wire's orientation and the field is a geometric property you control independently of $I$ or $L$. There's no mechanism here that rotates the wire. Choice B drops the $\sin30°$ factor entirely, as if the angle doesn't matter. This is a classic mistake — the angle always matters unless the wire is perpendicular to the field ($\sin90° = 1$). Choice D substitutes $\cos30°$ for $\sin30°$, confusing which component is relevant. The cosine would give the component parallel to the field, which contributes zero force. Study tip: Whenever you see $F = ILB$, immediately ask yourself, "where's the $\sin\theta$?" Its absence is almost always a trap.
4. What is the net magnetic force on the entire rectangular loop?
   Answer: $\vec{F}_{\text{net}} = 0$, because for any closed current loop in a uniform magnetic field, the forces on opposite sides are equal in magnitude and opposite in direction, so all contributions cancel regardless of the current direction or loop orientation. Whenever you see a question about the net force on a current loop in a uniform magnetic field, the single most important principle to recall is this: the net magnetic force on any closed current loop in a uniform field is always zero. Here's why. The net force on a current-carrying conductor is $\vec{F} = I\int d\vec{\ell} \times \vec{B}$. For a closed loop, you're integrating $d\vec{\ell}$ around a complete circuit. Since $\vec{B}$ is uniform (constant), it factors out: $\vec{F}_{\text{net}} = I\left(\oint d\vec{\ell}\right) \times \vec{B}$. The closed-loop integral $\oint d\vec{\ell} = \vec{0}$ because every displacement around the loop brings you back to the starting point — the vector sum of all path segments is zero. Therefore $\vec{F}_{\text{net}} = \vec{0}$, confirming D is correct. This holds regardless of loop shape, current direction, or orientation. Choice A is wrong because it claims forces on vertical segments only "partially cancel" — they cancel completely, and horizontal segments (parallel to $\vec{B}$) contribute zero force, not some net $\hat{y}$ force. Choice B is wrong because while $\hat{y} \times \hat{x} = -\hat{z}$ and $(-\hat{y}) \times \hat{x} = +\hat{z}$ are correct cross products, these forces occur on opposite sides of the loop and are equal and opposite — they cancel exactly. Choice C doubles down on B's error, incorrectly adding both vertical-side forces in the same direction rather than recognizing they oppose each other. Study tip: Always distinguish between net force (zero in a uniform field) and net torque (which can be nonzero). Exam questions often test whether you conflate these two — a uniform field can rotate a loop without translating it.
5. Which of the following statements best describes the net magnetic force on the closed loop?
   Answer: The net force is generally nonzero and directed in the $\pm x$-direction, because the field strength increases with $x$, so portions of the loop at larger $x$ experience a stronger force than portions at smaller $x$, and these unequal forces do not cancel for a non-uniform field. When a current loop sits in a non-uniform magnetic field, the key question is whether forces on opposite segments cancel. Start by writing the infinitesimal force element: $d\vec{F} = I\,d\vec{\ell} \times \vec{B}$. For this field, $\vec{B} = B_0(1+\alpha x)\hat{z}$, and since the loop lies in the $xy$-plane, every $d\vec{\ell}$ has only $\hat{x}$ and $\hat{y}$ components. Crossing those with $\hat{z}$ produces forces in the $xy$-plane — specifically, segments with $\hat{x}$ components produce $\hat{y}$ forces and vice versa. Crucially, because the field magnitude grows with $x$, segments on the right side of the loop are pulled with greater force than corresponding segments on the left. These forces don't cancel, producing a **net force in the $+x$-direction, confirming B** is correct. A is wrong because the identity $\oint d\vec{\ell} = 0$ only guarantees zero net force in a uniform field (where you can factor $\vec{B}$ outside the integral). When $\vec{B}$ varies with position, you cannot factor it out, so the integral does not simplify to zero. C is wrong on two counts: the net force on a planar loop in this field lies in the $xy$-plane, not $\hat{z}$, and the dipole-force formula $\vec{F} = \nabla(\vec{m}\cdot\vec{B})$ does not say the force is always along $\vec{B}$ itself. D is wrong because Green's theorem does not guarantee cancellation of force integrals — that theorem relates line integrals to area integrals of curls of scalar/vector functions, not to force balance in a position-dependent field. Study tip: On non-uniform field problems, immediately ask yourself: *can I factor $\vec{B}$ outside the integral?* If not, forces on opposite segments won't cancel, and you should expect a net force directed along the gradient of field strength.

Start this quiz