A 4.0 cm long straight wire carrying a current of 8A is placed perpendicular to an uniform magnetic field of strength 0.15 T. The magnetic force on the wire is ______ mN.
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Correct Answer: 48
Approach Solution - 1
The problem asks to calculate the magnetic force experienced by a straight current-carrying wire when it is placed perpendicular to a uniform magnetic field.
Concept Used:
The magnetic force (\( F_m \)) on a straight wire of length \( L \) carrying a current \( I \) in a uniform magnetic field of strength \( B \) is given by the formula:
\[ F_m = I L B \sin\theta \]where \( \theta \) is the angle between the direction of the current (along the length of the wire) and the direction of the magnetic field.
Step-by-Step Solution:
Step 1: List the given quantities and convert them to SI units.
The given values are:
- Length of the wire, \( L = 4.0 \, \text{cm} = 4.0 \times 10^{-2} \, \text{m} \)
- Current in the wire, \( I = 8 \, \text{A} \)
- Magnetic field strength, \( B = 0.15 \, \text{T} \)
The wire is placed perpendicular to the magnetic field, which means the angle \( \theta \) is \( 90^\circ \).
Step 2: Substitute the values into the magnetic force formula.
The formula for the magnitude of the magnetic force is:
\[ F_m = I L B \sin\theta \]Substituting the given values:
\[ F_m = (8 \, \text{A}) \times (4.0 \times 10^{-2} \, \text{m}) \times (0.15 \, \text{T}) \times \sin(90^\circ) \]Step 3: Calculate the magnetic force in Newtons.
Since \( \sin(90^\circ) = 1 \), the expression simplifies to:
\[ F_m = 8 \times 4.0 \times 10^{-2} \times 0.15 \] \[ F_m = 8 \times 0.04 \times 0.15 \] \[ F_m = 0.32 \times 0.15 \] \[ F_m = 0.048 \, \text{N} \]Final Computation & Result:
The problem asks for the magnetic force in milliNewtons (mN). To convert from Newtons (N) to milliNewtons (mN), we use the conversion factor \( 1 \, \text{N} = 1000 \, \text{mN} \).
\[ F_m = 0.048 \, \text{N} \times \frac{1000 \, \text{mN}}{1 \, \text{N}} = 48 \, \text{mN} \]The magnetic force on the wire is 48 mN.
Approach Solution -2
$F = IlB$
$F = 8 \times \frac{4}{100} \times 0.15$
$F = \frac{48 \times 100}{10000} N$
$F = 48 \times 10^{-3} N$
$F = 48 \text{ mN}$
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