## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 19 - Problems - Page 755: 52

#### Answer

(a) The magnetic force on the left side of the wire is $0.50~A$ (directed out of the page) The magnetic force on the top side of the wire is 0. The magnetic force on the right side of the wire is $0.50~A$ (directed into the page) The magnetic force on the bottom side of the wire is 0. (b) The net magnetic force on the loop is zero.

#### Work Step by Step

(a) We can find the magnetic force on the left side of the wire: $F = ILB$ $F = (1.0~A)(0.20~m)(2.5~T)$ $F = 0.50~A$ (By the right hand rule, this force is directed out of the page.) We can find the magnetic force on the top side of the wire: $F = ILB~sin~\theta$ $F = (1.0~A)(0.30~m)(2.5~T)~sin(180^{\circ})$ $F = 0$ We can find the magnetic force on the right side of the wire: $F = ILB$ $F = (1.0~A)(0.20~m)(2.5~T)$ $F = 0.50~A$ (By the right hand rule, this force is directed into the page.) We can find the magnetic force on the bottom side of the wire: $F = ILB~sin~\theta$ $F = (1.0~A)(0.30~m)(2.5~T)~sin(0^{\circ})$ $F = 0$ (b) Since the forces into the page and out of the page are equal in magnitude, the net magnetic force on the loop is zero.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.