Question

The two springs in the figure each have a spring constant of 11N/m. They are stretched by 0.60 cm when a current passes through the wire.

How big is the current?

Answer 1

Spring constant (k) of each spring = 11 N/m

Stretched distance (x) of each spring = 0.60 cm = 0.006 m

Here the springs are connected in the parallel combination, so the effective spring consant is

                     k_eff = k₁ + k₂

                           = k + k

                           = 2k

                           = 2(11 N/m)

                           = 22 N/m

Therefore, the force acting on the spring is balanced with force acting on the current carrying wire due to the magnetic field.

Therefore, k_effx = BIL

                      I = (k_effx) / (BL)

                        = [(22 N/m)(0.006 m)] / [(0.50 T)(20 cm)]

                        = [(22 N/m)(0.006 m)] / [(0.50 T)(0.20 m)]

                        = 1.32 A --> Answer

Answer 2

Firstly we must calculate the amount of force needed to produce a stretch of 0.60 cm:

F = k*d where F is Force (N), k is the Spring Constant (N/m) and d is the Dispacement from its original position. (m)

F = 11 * 0.60x10^-2 = 6.6x10^-2 N

This force is present as the current is at 90 degrees to the magnetic field. We can use the following formula to calculate the current through the wire:

Current = F / B*l where B is the Magnetic Field Strength (T) and l is the Length of the wire in the Magnetic Field (m)

Current = 6.6x10^-2 / ( 0.50 * 0.20 ) = 0.66A