A 4.0 cm long straight wire carrying a current of 8A is placed perpendicular to a uniform magnetic field of strength 0.15 T. The magnetic force on the wire is __________ mN.
Magnetic Force Calculation on a Wire
The magnetic force ($F$) exerted on a straight wire carrying current ($I$) placed in a uniform magnetic field ($B$) is calculated using the formula:
$ F = I \times L \times B \times \sin(\theta) $
Where:
- $I$ is the current in Amperes (A).
- $L$ is the length of the wire in meters (m).
- $B$ is the magnetic field strength in Tesla (T).
- $\theta$ is the angle between the direction of the current and the magnetic field.
Given Values
From the question, we have:
- Length of the wire, $L = 4.0 \text{ cm}$. We need to convert this to meters: $L = 4.0 / 100 = 0.04 \text{ m}$.
- Current, $I = 8 \text{ A}$.
- Magnetic field strength, $B = 0.15 \text{ T}$.
- The wire is placed perpendicular to the magnetic field, so $\theta = 90^\circ$.
- We know that $\sin(90^\circ) = 1$.
Applying the Formula
Substitute the values into the formula:
$ F = (8 \text{ A}) \times (0.04 \text{ m}) \times (0.15 \text{ T}) \times \sin(90^\circ) $
$ F = 8 \times 0.04 \times 0.15 \times 1 \text{ N} $
Calculating the Force
First, multiply the current and length:
$ 8 \times 0.04 = 0.32 $
Now, multiply the result by the magnetic field strength:
$ F = 0.32 \times 0.15 \text{ N} $
$ F = 0.048 \text{ N} $
Converting to MilliNewtons (mN)
The question asks for the force in milliNewtons (mN). To convert Newtons (N) to milliNewtons (mN), we multiply by 1000:
$ F = 0.048 \times 1000 \text{ mN} $
$ F = 48 \text{ mN} $
Final Answer
The magnetic force on the wire is 48 mN.
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