I need help with 2 and 3. 1. What is the force due to the Earth’s magnetic field

Question

I need help with 2 and 3.

1. What is the force due to the Earth’s magnetic field (magnitude 0.00005 T) on a horizontal wire 12 cm long carrying one amp of current? Assume the Earth’s magnetic field is perpendicular to the wire.

2. Assume the mass of the wire described in problem 1 is 2 grams. Place another wire parallel to this wire, at a distance of 2 mm below it. How much current must flow in this second wire to lift the first wire? Also, what direction must the current flow in the second wire?

3 Extra Credit: Determine for problem 1 how much power is dissipated by the two-gram wire, if it is made of copper. The density of copper is 9 grams per cubic centimeter. The resistivity ? of copper is 1.7 times 10?8?m.

Explanation / Answer

1)

magneitc force F = i*L*B*sintheta

current i =1 A

length L = 12 cm = 0.12 m

magnetic field B = 0.00005 T

theta = angle between L and B = 90

magnetic force = 1*0.12*0.00005*sin90 = 6*10^-6 N

(2)

there should be force of repulsion between 1 and 2

force on wire 1 = weight of wire 1

magnetic force on wire1 = uo*I1*I2*L1/(2*pi*r)

weight = m*g

uo*I1*I2*L1/(2*pi*r) = m*g

I2 = m*g*2*pi*r/(uo*I1*L)

I2 = (2*10^-3*9.8*2*pi*2*10^-3)/(4*pi*10^-7*1*0.12)

I2 = 1633.33 A

======================

3)

power dissipated = I1^2*R

R = resistance = rho*L/A*(L/L) = rho*L^2/V

L = length of wire = 0.12 m

V = volume = mass/density = m/D

rho = resistivity

Resistance = rho*D*L^2/m

Resistane = 1.7*10^-8*9*10^3*0.12^2/(2*10^-3) = 0.0011 N

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