Question

A square loop made of copper wire is falling out of a region where there is a magnetice field B_ext = 0.06 T. The loop is falling with a constant velocity because of the current induced in the loop.

What is the velocity of the loop? Give your answer in m/s to at least three significant digits to avoid being counted wrong for rounding.

It will be useful to know that:

The mass density of copper (mass per volume) is 9.0x10³ kg/m³.

The volume of the wire is A L, where A is the cross-sectional area of the wire and L is the total length (sum of the length of the four sides).

The resistivity of copper is ρ = 1.7x10^-8 Ω-m.

The resistance of the wire is R = ρ L/A, where A and L are again the cross-sectional area and length of the wire

Answer 1

Let I is the current through loop,

here magnetic force = gravittional force

i. e i L B = m g

B*I*L = m*g

I = m*g/(B*L) ----------------------------------(1)

we know, induced emf = B*v*L

induced current, I = induced emf/R

Current I = B*v*L/R ---------------------------------(2)

Solving 1 and 2

B*v*L/R = m*g/(B*L)

v = m*g*R/(B^2*L^2)

multiply and deivide with area A

v = m*g*R*A/(B^2*L^2*A)

= m/(L*A) * (R*A/L) *(1/B^2)

= 4*m/(4*L*A) * (R*A/L) *(1/B^2)

Speed v = 4*density*resistivity/B^2 is the formula for sped v

so Now Upon Substituin

v = 4* 9000 * 1.7 *10^-8/(0.06^2)

v = 0.17 m/s