Question

The direction of the force on a current-carrying wire in an external magnetic field is
a
perpendicular to the current only.
b
parallel to the current and to the magnetic field.
c
perpendicular to the magnetic field only.
d
perpendicular to the current and to the magnetic field.
Question 15 (1 point)
A 2.0 m wire segment carrying a current of 0.60 A oriented parallel to a uniform magnetic field of 0.50 T experiences a force of what magnitude?
a
0.0 N
b
0.15 N
c
0.30 N
d
0.60 N

Answer 1

The direction of the force on a current-carrying wire in an external magnetic field is given by the right-hand rule and is determined by the equation for the magnetic force:

\[ F = I L B \sin(\theta) \]

where:
- \( F \) is the force,
- \( I \) is the current (in Amperes),
- \( L \) is the length of the wire (in meters),
- \( B \) is the magnetic field strength (in Teslas), and
- \( \theta \) is the angle between the direction of the current and the magnetic field.

From this equation, we can see that the force is maximum when the current and magnetic field are perpendicular (i.e., \( \theta = 90^\circ \)). Thus, the correct answer regarding the force's direction is:

**d) perpendicular to the current and to the magnetic field.**

For the second question, since the wire is oriented parallel to the magnetic field, \(\theta = 0^\circ\), and thus \(\sin(0) = 0\). Therefore, using the equation for force:

\[ F = I L B \sin(0) = 0 \]

As a result, the force experienced by the wire segment is:

**a) 0.0 N**.