# Chapter 19 - Problems - Page 755: 46

(a) The magnitude of the maximum possible force is $2.16~N$ (b) The information enables us to calculate the maximum possible force in the case when the angle $\theta$ is $90^{\circ}$. However, the force could be less than this magnitude depending on the angle $\theta$.

#### Work Step by Step

(a) We can find the magnitude of the maximum possible force: $F = I~L~B~sin~\theta$ $F = (18.0~A)(0.60~m)(0.20~T)~sin~\theta$ $F = (2.16~sin~\theta)~N$ The magnitude of the maximum possible force is $2.16~N$ (b) The information enables us to calculate the maximum possible force in the case when the angle $\theta$ is $90^{\circ}$. However, the force could be less than this magnitude depending on the angle $\theta$.

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