A horizontal wire of length 0.50 m and weight 1.0 N is placed in a uniform horizontal magnetic field of flux density 1.5 T directed at 90° to the wire - AQA - A-Level Physics - Question 24 - 2019 - Paper 2

The magnetic force ( ( F_B )) on a current-carrying wire in a magnetic field can be calculated using the formula: $F_B = BIL$ where:

• ( B ) is the magnetic flux density (1.5 T),
• ( I ) is the current in the wire,
• ( L ) is the length of the wire (0.50 m).

We need to find the current that will produce a magnetic force equal to the weight of the wire (1.0 N).

To find the current that supports the wire, we set the magnetic force equal to the weight: $F_B = W$ So, $BIL = W$ Substituting the values: $1.5 I (0.50) = 1.0$

Rearranging the equation gives us: $I = \frac{1.0}{1.5 \times 0.50}$ $I = \frac{1.0}{0.75} = 1.33 A$ Thus, the current that just supports the wire is approximately 1.33 A.

From the given options, the closest value is 1.3 A (Option C).