Physics 2 Quiz: Magnetic Force On Wire

What this quiz covers

This quiz focuses on Magnetic Force On Wire, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $0.80 \text{ N}$, directed in the $+y$-direction, because only the $z$-component of $\vec{B}$ contributes to the cross product $\hat{x} \times \hat{z}$, yielding $F = ILB_z = (4.0)(0.50)(0.40)$. (correct answer)
2. $0.70 \text{ N}$, because both field components contribute equally and the net force magnitude is $ILB_{\text{total}} = (4.0)(0.50)(0.50)$, where $B_{\text{total}} = \sqrt{0.30^2 + 0.40^2}$.
3. $0.60 \text{ N}$, because the $x$-component of $\vec{B}$ is parallel to the current and therefore contributes a force $ILB_x = (4.0)(0.50)(0.30)$, which must be added to the force from the $z$-component.
4. $0.28 \text{ N}$, because the $x$-component of $\vec{B}$ is parallel to the current and contributes nothing, while the $z$-component contributes $ILB_x\sin(\theta)$ where $\theta$ is the angle between the two non-zero components.

Question 2

1. $|\vec{F}| = ILB_0/\sqrt{2} \approx 0.85 \text{ N}$, directed in the $-\hat{z}$-direction, because the $\hat{y}$-component of the wire is parallel to $\vec{B}$ and produces no force, and the $\hat{x}$-component gives $\hat{x}\times\hat{y}$; applying the right-hand rule with the fingers curling from $\hat{x}$ toward $\hat{y}$ yields a force in $-\hat{z}$.
2. $|\vec{F}| = ILB_0/\sqrt{2} \approx 0.85 \text{ N}$, 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 $\hat{y}\times\hat{y}=0$. (correct answer)
3. $|\vec{F}| = ILB_0 = (5.0)(0.40)(0.60) = 1.2 \text{ N}$, directed in the $+\hat{z}$-direction, because the magnetic force formula $F=ILB$ uses the full length of the wire and the full field magnitude regardless of the wire's orientation, since every element of the wire is immersed in the field $\vec{B}$.
4. $|\vec{F}| = ILB_0\cos45° \approx 0.85 \text{ N}$, directed in the $+\hat{y}$-direction, because the component of the wire along $\hat{y}$ (equal to $L/\sqrt{2}$) is parallel to $\vec{B}$ and therefore experiences the strongest interaction, producing a force in the $+\hat{y}$-direction, while the perpendicular component's contribution is negligible.

Question 3

1. $\vec{F}_{\text{net}} = 1.2\hat{y} \text{ N}$, because the horizontal segments of the loop carry current along $\hat{x}$, which is parallel to the field, while the vertical segments produce forces that partially cancel, leaving a net force in $\hat{y}$.
2. $\vec{F}_{\text{net}} = 1.2\hat{z} \text{ N}$, because the two vertical sides of the loop both carry current with a $y$-component, and $\hat{y} \times \hat{x} = -\hat{z}$ for one side and $(-\hat{y}) \times \hat{x} = +\hat{z}$ for the other, but these forces act on the same side and do not cancel.
3. $\vec{F}_{\text{net}} = 2.4\hat{z} \text{ N}$, because the net force is computed by summing only the forces on the two sides parallel to $\hat{y}$, each contributing $IbB = (5.0)(0.40)(0.60) = 1.2 \text{ N}$ in the same $\hat{z}$-direction.
4. $\vec{F}_{\text{net}} = 0$, 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. (correct answer)

Question 4

1. The net force is zero, because the magnetic force on any closed current loop is always zero regardless of whether the field is uniform or non-uniform, as a direct consequence of the closed-path integral $\oint d\vec{\ell} = 0$.
2. 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. (correct answer)
3. The net force is nonzero and directed in the $+z$-direction, because the curl of $\vec{B}$ is nonzero ($\partial B_z/\partial x = B_0\alpha \neq 0$), and the net force on a magnetic dipole in a non-uniform field is always directed along the field vector $\vec{B}$, which points in $+\hat{z}$.
4. The net force is zero, because although the field is non-uniform, every infinitesimal force element $Id\vec{\ell}\times\vec{B}$ lies in the $xy$-plane, and integrating these in-plane forces around any closed loop always yields zero by Green's theorem applied to planar vector fields.

Question 5

1. The claim is incorrect, and the original force is $F = ILB\sin30° = 0.30 \text{ N}$, because doubling $I$ and halving $L$ preserves $IL$ only if both changes are exact, but the resulting wire geometry changes the effective angle of force, altering $\sin\theta$ and therefore the net force.
2. The claim is correct, and the original force is $F = ILB = (8.0)(0.30)(0.25) = 0.60 \text{ N}$, because the $30°$ angle does not modify the force formula when the wire is not perpendicular to the field, and $IL$ is preserved under the proposed changes.
3. 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 $2.4 \text{ A}\cdot\text{m}$. (correct answer)
4. The claim is incorrect, and the original force is $F = ILB\cos30° = (8.0)(0.30)(0.25)(0.866) = 0.52 \text{ N}$, because the relevant projection of the wire onto the field direction uses $\cos30°$, and shortening the wire changes this projection in a way that does not preserve the force.