Question:

A wire carrying a current of 2 A is placed in a magnetic field. The wire is oriented at a 30° angle with respect to the magnetic field lines. The wire experiences a magnetic force of 0.5 N. Determine the magnitude of the magnetic field strength.

(a) 0.25 T
(b) 0.43 T
(c) 1.15 T
(d) 1.73 T

Answer:

The magnitude of force experienced by a current-carrying wire in a magnetic field can be calculated using the formula:

Where:

• $F$ is the force experienced by the wire (N),
• $B$ is the magnetic field strength (T),
• $I$ is the current flowing through the wire (A),
• $L$ is the length of the wire in the magnetic field (m), and
• $\theta$ is the angle between the wire and the magnetic field lines.

In this question, we are given:

• $I = 2$ A,
• $\theta = 30^\circ$, and
• $F = 0.5$ N.

We need to solve for $B$.

Rearranging the formula, we have:

Plugging in the values, we get:

Since the length of the wire is not specified in the question, we can assume a length of 1 meter without loss of generality. Therefore, $L = 1$ m.

Simplifying, we get:

Therefore, the magnitude of the magnetic field strength is 0.5 T.