A copper rod of length 0.83 m is lying on a frictionless table (see the drawing). Each end of the rod is attached to a fixed wire by an unstretched spring whose spring constant is k = 79 N/m. A magnetic field with a strength of 0.13 T is oriented perpendicular to the surface of the table. If the current is 18 A, by how much does each spring stretch?

To determine how much each spring stretches, we need to consider the forces acting on the copper rod.

Let's start by calculating the magnetic force acting on the rod. The magnetic force on a current-carrying wire in a magnetic field can be found using the equation:

F = I * L * B * sin(theta)

Where:
F is the magnetic force
I is the current
L is the length of the wire
B is the magnetic field strength
theta is the angle between the wire and the magnetic field (in this case, it's 90 degrees)

Plugging in the given values:
I = 18 A (current)
L = 0.83 m (length of the rod)
B = 0.13 T (magnetic field strength)
theta = 90 degrees

F = 18 A * 0.83 m * 0.13 T * sin(90)

Since sin(90) = 1,
F = 18 A * 0.83 m * 0.13 T

Now, let's consider the forces exerted by the two springs. Each spring obeys Hooke's Law, which states that the force exerted by a spring is proportional to its deformation:

F = -k * x

Where:
F is the force exerted by the spring
k is the spring constant
x is the deformation (stretching or compression) of the spring from its equilibrium position

Since we have two springs, the total force exerted by the springs is equal to the sum of the individual forces:

F_total = -k * x1 - k * x2

Since the forces exerted by the springs are equal in magnitude, we can simplify it to:

F_total = -2 * k * x

Now, we can equate the magnetic force and the spring force:

-2 * k * x = 18 A * 0.83 m * 0.13 T

Simplifying the equation:
x = (18 A * 0.83 m * 0.13 T) / (2 * k)

Plugging in the values:
k = 79 N/m (spring constant)

x = (18 A * 0.83 m * 0.13 T) / (2 * 79 N/m)

Now we can calculate x:

x ≈ 0.169 meters

Therefore, each spring stretches by approximately 0.169 meters.