## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 19 - Problems - Page 755: 51

#### Answer

(a) The magnetic force on the left side of the wire is $0.50~A$ (directed to the right) The magnetic force on the top side of the wire is $0.75~A$ (directed downward) The magnetic force on the right side of the wire is $0.50~A$ (directed to the left) The magnetic force on the bottom side of the wire is $0.75~A$ (directed upward) (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 to the right.) We can find the magnetic force on the top side of the wire: $F = ILB$ $F = (1.0~A)(0.30~m)(2.5~T)$ $F = 0.75~A$ (By the right hand rule, this force is directed downward.) 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 to the left.) We can find the magnetic force on the bottom side of the wire: $F = ILB$ $F = (1.0~A)(0.30~m)(2.5~T)$ $F = 0.75~A$ (By the right hand rule, this force is directed upward.) (b) Since the forces to the left and right are equal in magnitude, and the forces downward and upward are equal in magnitude, the net magnetic force on the loop is zero.

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