# An electromagnet can exert a lifting force of 12.0N when it operates from a current of 5.0A. What mass can it lift if the current is increased to 8.0A?

May 12, 2017

The electromagnet can lift a mass of 3.1 kg.

#### Explanation:

Calculate the lifting force

The equation for the lifting force of an electromagnet is

color(blue)(bar(ul(|color(white)(a/a) F = ((nI)^2µ_0A)/(2g^2)color(white)(a/a)|)))" "

where

$n =$ the number of terms
$I =$ the current
µ_0 =  the permeability of air
$A =$ the area of the poles
$g =$the gap between the object and the electromagnet

In this problem, you are changing only the current, so you can write

F_2/F_1 = ((color(red)(cancel(color(black)(n)))I_2)^2color(red)(cancel(color(black)(µ_0A))))/((color(red)(cancel(color(black)(n)))I_1)^2color(red)(cancel(color(black)(µ_0A)))) × stackrelcolor(blue)(1)(cancel(color(black)2g^2))/(color(red)(cancel(color(black)(2g^2))))

${F}_{2} / {F}_{1} = {\left({I}_{2} / {I}_{1}\right)}^{2}$

F_2 = F_1 × (I_2/I_1)^2

In this problem.

${F}_{1} = \text{12.0 N"; I_1 = "5.0 A}$
F_2 = ?;color(white)(mml) I_2 = "8.0 A"

F_2 = "12.0 N" × ((8.0 color(red)(cancel(color(black)("A"))))/(5.0 color(red)(cancel(color(black)("A")))))^2 = 30.7 N

Calculate the mass that can be lifted

$F = m g$

M = F/g = (30.7 color(red)(cancel(color(black)("N"))))/(9.81 color(red)(cancel(color(black)("m·s"^"-2")))) × ("1 kg"·color(red)(cancel(color(black)("m·s"^"-2"))))/(1 color(red)(cancel(color(black)("N")))) = "3.1 kg"

The electromagnet can lift a mass of 3.1 kg.