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(B) The magnetic field is given by $\vec{B} = B_0(1+4x) \hat{k}$.For the vertical wire at $x=0$,the magnetic field is $B(0) = B_0(1+4(0)) = B_0 = 5 \

Vedclass · Accepted Answer

$160$

Vedclass · Answer

(B) The magnetic field is given by $\vec{B} = B_0(1+4x) \hat{k}$.For the vertical wire at $x=0$,the magnetic field is $B(0) = B_0(1+4(0)) = B_0 = 5 \ T$.The force on this wire is $F_1 = i \ell B(0) = 2 	imes 2 	imes 5 = 20 \ N$ (directed in the $+x$ direction).For the vertical wire at $x=2$,the magnetic field is $B(2) = B_0(1+4(2)) = 9B_0 = 9 	imes 5 = 45 \ T$.The force on this wire is $F_2 = i \ell B(2) = 2 	imes 2 	imes 45 = 180 \ N$ (directed in the $-x$ direction).The forces on the horizontal wires cancel each other out.The net force is $F_{net} = F_2 - F_1 = 180 - 20 = 160 \ N$.

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