Question

What force is exerted on the water in an MHD drive utilizing a 25.0-cm-diameter tube, if 100-A current is passed across the tube that is perpendicular to a 2.00-T magnetic field? (The relatively small size of this force indicates the need for very large currents and magnetic fields to make practical MHD drives.)

a) 0.5 N
b) 5.0 N
c) 50.0 N
d) 500.0 N

Answer 1

The force exerted on water within a 25.0-cm-diameter tube in an MHD drive, with a 100-A current and a 2.00-T magnetic field perpendicular to the current, is calculated using the equation F = IBL sin(theta) and comes out to be 50.0 N.

Step-by-step explanation:

To calculate the force exerted in a magnetohydrodynamic (MHD) drive, we use the equation F = IBL sin(θ), where I is the current, B is the magnetic field strength, L is the length of the conductor (tube) in the magnetic field, and θ is the angle between the direction of the current and the magnetic field. With the given current of 100 A, a magnetic field strength of 2.00 T, and the tube diameter of 25.0 cm (which gives us the circumference as the length of the conductor in the field), we can calculate the force. Given that the current is perpendicular to the magnetic field, θ is 90 degrees, making sin(θ) = sin(90°) = 1.

The force exerted on the water in the tube can be calculated as follows:

• Calculate the length of the conductor (tube) in the field, L = π x diameter = 3.14159 x 0.25 m.
• Calculate the force, F = I x B x L x sin(90°) = 100 A x 2.00 T x (3.14159 x 0.25 m) x 1.
• After calculation, the force, F, comes out to be approximately 50.0 N, making the correct answer (c) 50.0 N.