Question

In a certain region of space near earth's surface, a uniform horizontal magnetic field of magnitude B exists above a level defined to be y = 0. Below y = 0 , the field abruptly becomes zero (seethe figure). A vertical square wire loop has resistivity rho mass density rhom, diameter d, and side length l. It is initially at rest with its lower horizontal side at y = 0 and is then allowed to fall under gravity, with its plane perpendicular to the direction of the magnetic field.
a) While the loop is still partially immersed in the magnetic field (as it fallsinto the zero-field region), determine the magnetic "drag" forcethat acts on it at the moment when its speed is v.
b) Assume that the loop achieves a terminal velocity vt before its upper horizontal side exits the field. Determine a formulafor vt
c) If the loop is made of copper and B = 0.80 T find vt

Answer 1

Answer:

a) F = $\frac{\pi d^2B^2lv}{16p}$  

b) attached below

c) 0.037 m/s

Explanation:

a) Determine the magnetic "drag" force acting  at the moment

speed = v

first step: determine current in the loop

I = $\frac{\pi d^2}{16pl} B lv$   ----- ( 1 )

given that the current will induce force on the three sides of the loop found in the magnetic field

forces on vertical sides = + opposite

we will cancel out

hence equation 1 becomes

F = $\frac{\pi d^2B^2lv}{16p}$   ( according to Lenz law we can say that the direction of force is upwards and this force will slow down the decrease in flux )

b) Determine the formula for Vt

attached below

c) Find Vt

given :

B = 0.80 T

density of copper = 8.9 * 10^3 kg/m^3

resistivity of copper = 1.68 * 10^-8 Ωm

∴ Vt = 16 ( 8.9 * 10^3 kg/m^3 ) ( 1.68 * 10^-8 Ωm ) ( 9.8 m/s^2 ) / ( 0.08 T)^2

       = 0.037 m/s