Question

learning goal: to understand the magnetic force on a straight current - carrying wire in a uniform magnetic field. part a consider a wire of length (l = 0.30 m) that runs north - south on a horizontal surface. there is a current of (i = 0.50 a) flowing north in the wire (the rest of the circuit, which actually delivers this current, is not shown.) the earths magnetic field at this location has a magnitude of 0.50 gauss (or, in si units, (0.5\times10^{-4}) tesla) and points north and 38 degrees down from the horizontal, toward the ground. what is the size of the magnetic force on the wire due to the earths magnetic field? in considering the agreement of units, recall that (1 t=1 n/(acdot m)) (figure 1) express your answer in newtons to two significant figures. view available hint(s)

Explanation:

Step1: Identify the formula for magnetic - force

The formula for the magnetic force on a current - carrying wire is $F = ILB\sin\theta$, where $I$ is the current, $L$ is the length of the wire, $B$ is the magnetic field strength, and $\theta$ is the angle between the direction of the current and the magnetic field.

Step2: Determine the values of variables

Given $I = 0.50\ A$, $L=0.30\ m$, $B = 0.5\times10^{-4}\ T$, and the angle $\theta = 38^{\circ}$ (since the current is north - south and the magnetic field is north and 38 degrees down from the horizontal).

Step3: Calculate the magnetic force

Substitute the values into the formula: $F=ILB\sin\theta=(0.50\ A)\times(0.30\ m)\times(0.5\times 10^{-4}\ T)\times\sin(38^{\circ})$.
First, calculate $\sin(38^{\circ})\approx0.616$. Then $F=(0.50)\times(0.30)\times(0.5\times 10^{-4})\times0.616\ N$.
$F = 4.62\times10^{-6}\ N\approx4.6\times 10^{-6}\ N$.

Answer:

$4.6\times 10^{-6}\ N$