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 acting on the wire is approximately 0.0208 T (Tesla).

Step-by-step solution

List down the known values from the problem

The given values from the problem are: Length of the wire, \(L = 2.00 ~m\) Current in the wire, \(I = 24.0 ~A\) Angle between the wire and magnetic field, \(\theta = 30.0^{\circ}\) Force on the wire, \(F = 0.500 ~N\)

Convert the angle to radians

We need to convert the angle from degrees to radians. To do so, we use the formula: \(angle (radians) = angle (degrees) \times \frac{\pi}{180}\) So, \(\theta = 30.0^{\circ} \times \frac{\pi}{180} = \frac{\pi}{6} ~radians\)

Rearrange the magnetic force formula to solve for B

We need to determine the magnitude of the magnetic field, B. Rearrange the magnetic force formula, \(F = I * L * B * \sin{\theta}\), to solve for \(B\): \(B = \frac{F}{I * L * \sin{\theta}}\)

Plug in the known values and calculate B

Now plug in the known values into the rearranged formula for \(B\): \(B = \frac{0.500 ~N}{24.0 ~A * 2.00 ~m * \sin{\frac{\pi}{6}}}\)

Calculate the magnetic field B

Compute the value for \(B\): \(B = \frac{0.500 ~N}{24.0 ~A * 2.00 ~m * 0.5} = \frac{0.500 ~N}{24.0 ~A} = 0.0208~T\) The magnitude of the magnetic field is approximately \(0.0208 ~T\) (Tesla).