jmc

Squaring the equation $a+b+c=0,$ we get $$a^2+b^2+c^2+2(ab+ac+bc)=4+2(ab+ac+bc)=0,$$so $ab+ac+bc=-2.$

Squaring the equation $a^2+b^2+c^2=4,$ we get $$a^4+b^4+c^4+2(a^2{b}^2+a^2{c}^2+b^2{c}^2)=16.$$Squaring the equation $ab+ac+bc=-2,$ we get $$a^2{b}^2+a^2{c}^2+b^2{c}^2+2abc(a+b+c)=4,$$so $a^2{b}^2+a^2{c}^2+b^2{c}^2=4.$

Therefore, $$a^4+b^4+c^4=16-2(a^2{b}^2+a^2{c}^2+b^2{c}^2)=16-2\cdot 4=\boxed{8}.$$