A 20.0-cm length of wire is aligned along the x axis and carries 5.0 mA of current. The wire is in a region where the uniform magnetic field is given by 1.5(i + 3j + 4k) mT. The net force on the wire is
A) 1.5 $\mu$ N.
B) 3.0$\mu$ N.
C) 4.5 $\mu$ N.
D) 7.5 $\mu$ N.

Question

Quizplus · Accepted Answer

The net force on the wire can be calculated using the formula for the magnetic force on a current-carrying wire: $F = I(L \times B)$, where $I$ is the current, $L$ is the length vector of the wire, and $B$ is the magnetic field vector. Given that the wire is aligned along the x-axis, its length vector can be represented as $L = 0.20\hat{i}$ m (since 20.0 cm = 0.20 m), the current $I = 5.0 \times 10^{-3}$ A (since 5.0 mA = 5.0 x 10^{-3} A), and the magnetic field $B = 1.5\hat{i} + 4.5\hat{j} + 6.0\hat{k}$ mT = $1.5 \times 10^{-3}\hat{i} + 4.5 \times 10^{-3}\hat{j} + 6.0 \times 10^{-3}\hat{k}$ T (since 1 mT = 10^{-3} T). The cross product $L \times B$ gives a vector that is perpendicular to both $L$ and $B$, and its magnitude can be calculated as the determinant of a matrix formed by the unit vectors $\hat{i}, \hat{j}, \hat{k}$ and the components of $L$ and $B$. However, since $L$ has only an $\hat{i}$ component, the cross product simplifies to $L \times B = 0.20(0\hat{i} - 6.0\hat{j} + 4.5\hat{k}) = -1.2\hat{j} + 0.9\hat{k}$. The magnitude of this vector is $\sqrt{(-1.2)^2 + (0.9)^2}$ = $\sqrt{1.44 + 0.81}$ = $\sqrt{2.25}$ = 1.5 m. Therefore, the magnitude of the force is $F = I \times 1.5 = 5.0 \times 10^{-3} \times 1.5 = 7.5 \times 10^{-3}$ N, which corresponds to 7.5 mN.

A 200-Cm Length of Wire Is Aligned Along the X Axis