Question:

Consider a straight wire carrying a current of 2.5 A. The wire is placed in a uniform magnetic field of 0.75 T directed at a 45° angle with respect to the wire. Determine the force acting on a 0.25 m length of wire.

Answer:

Given: Current (I) = 2.5 A Magnetic field (B) = 0.75 T Length of wire (L) = 0.25 m Angle between wire and magnetic field (θ) = 45°

To find the force acting on a length of wire, we will use the formula:

F = BILsin(θ)

where: F = Force acting on the wire (in Newtons) B = Magnetic field strength I = Current L = Length of wire θ = Angle between the wire and the magnetic field

Plugging in the given values, we have:

F = (0.75 T) x (2.5 A) x (0.25 m) x sin(45°)

Before proceeding, let's calculate the value of sin(45°):

sin(45°) = 0.707 (rounded to three decimal places)

Now, let's substitute the values into the formula:

F = (0.75 T) x (2.5 A) x (0.25 m) x 0.707

Simplifying the equation:

F = 0.132 J

Therefore, the force acting on the 0.25 m length of wire is 0.132 Newtons.

This force acts perpendicular to the wire and in the direction dictated by the right-hand rule (if the thumb points in the direction of the current, the fingers curl in the direction of the force).

Note: The direction of the force can also be determined using the cross product between the current and magnetic field vectors.