Question

A straight wire of length \(2.00 \mathrm{~m}\) carries a current of \(24.0 \mathrm{~A} .\) It is placed on a horizontal tabletop in a uniform horizontal magnetic field. The wire makes an angle of \(30.0^{\circ}\) with the magnetic field lines. If the magnitude of the force on the wire is \(0.500 \mathrm{~N}\), what is the magnitude of the magnetic field?

Short answer

Answer: The magnitude of the magnetic field is approximately 0.0208 T (tesla).

Step-by-step solution

Write down the known values

We are given the following values: - Length of the wire (\(L\)) = 2.00 m - Current in the wire (\(I\)) = 24.0 A - Angle between the wire and magnetic field lines (\(\theta\)) = 30.0° - Force on the wire (\(F\)) = 0.500 N

Convert the angle to radians

In order to use the formula, we need to convert the angle from degrees to radians. We can do this using the following conversion: 1 radian = 180°/π. So, we have: $$ \theta_{rad} = 30.0° \cdot \frac{\pi}{180°} = \frac{\pi}{6} \, \text{radians} $$

Use the formula

Now that we have all the necessary values, we can use the formula for the force on a wire in a magnetic field, \(F = BIL\sin{\theta}\), to find the magnitude of the magnetic field, \(B\): $$ B = \frac{F}{IL\sin{\theta}} $$

Substitute the known values

Now, substitute the known values into the formula: $$ B = \frac{0.500 N}{(24.0 A)(2.00 m) \sin(\frac{\pi}{6})} $$

Calculate the magnetic field magnitude

Finally, calculate the magnetic field magnitude: $$ B = \frac{0.500 N}{(24.0 A)(2.00 m) \, (0.5)} = \frac{0.500 N}{24.0 A} $$ The magnitude of the magnetic field is approximately: $$ B \approx 0.0208 \, \text{T} $$ So the magnitude of the magnetic field is approximately 0.0208 T (tesla).