A horizontal wire of length 0.25 m carrying a current of 3.0 A is perpendicular to a magnetic field - AQA - A-Level Physics - Question 22 - 2022 - Paper 2

Using the formula for the force on a current-carrying conductor in a magnetic field:

$F = BIL$

Where:

• $F = W = 2.943 \times 10^{-2} \, \text{N}$ (force equal to the weight of the wire)
• $I = 3.0 \, \text{A}$ (current)
• $L = 0.25 \, \text{m}$ (length of the wire)

Rearranging the formula to solve for the magnetic flux density ($B$):

$B = \frac{F}{IL} = \frac{2.943 \times 10^{-2} \, \text{N}}{3.0 \, \text{A} \times 0.25 \, \text{m}}$

Calculating:

$B \approx 3.9 \times 10^{-2} \, \text{T}$

The calculated magnetic flux density is approximately $3.9 \times 10^{-2} \, \text{T}$, which corresponds to option B.