Question

A rectangular loop of wire is placed next to a straight wire, as shown in the...

A rectangular loop of wire is placed next to a straight wire, as shown in the (Figure 1) . There is a current of

I = 7.0A in both wires.

Determine the magnitude of the net force on the loop

Physics


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Answer #1
Concepts and reason

The concepts used to solve this problem are magnetic field of a long wire and force acting on a wire carrying current.

Use the current in the wire and the distance from the wire relation to calculate the magnetic field of the long wire.

Use the current, length of the wire, magnetic field to find the force on wire at side ‘a’ and ‘c’ due to the current carrying wire.

Fundamentals

The figure depicts rectangular loop of wire is placed next to a straight line,

.0 cm
a
-
5.0 cm
d
10.0 cm
Figure 1

Expression for the magnetic field at side ‘a’ due to current carrying straight wire is,

2πη

Here, is the magnetic field at side ‘a’ due to current carrying straight wire, I is the current, is the permeability, and r is the distance between the current carrying wire and wire at ‘a’.

Expression for the force on wire at side ‘a’ due to current carrying wire is,

F. = IIB.

Here, is the force on wire at side ‘a’ due to current carrying wire and l is the length of the wire ‘a’.

Substitute μ, /2πη
for .

и?
2л

Expression for the magnetic field at side ‘c’ due to current carrying straight wire is,

B.
21R

Here, is the magnetic field at side ‘c’ due to current carrying straight wire and R is the distance between the current carrying wire and wire at ‘c’.

Expression for the force on wire at side ‘c’ due to current carrying wire is,

F = IIB

Here, is the force on wire at side ‘c’ due to current carrying wire and l is the length of the wire ‘c’.

Substitute 41/2nR
for .

41²1
20R

Expression for the net force on the loop is,

Foner = F,-F

Here, is the net force on the loop.

Expression for the force on wire at side ‘a’ due to current carrying wire is,

2ле

Substitute 47x10-N-A?
for , for , 10.0 cm
for , 3.00 cm
for .

(47x10-?N-A?)(7.0A)? (10.0cm)
27 (3.00cm)
= 3.266x10 N

Expression for the force on wire at side ‘c’ due to current carrying wire is,

21R

Substitute 47x10-N-A?
for , for , 10.0 cm
for , 8.00 cm
for .

_(47x10-N-A?)(7.04) (10.0cm)
21(8.00cm)
= 1.225x10 N

Expression for the net force on the loop is,

Foner = F,-F

Substitute 3.266x10 N
for and 1.225x10 N
for .

Fe = 3.266x10-N-1.225x10-SN
= 2.041x10 N
2.04x10- N

Ans:

The magnitude of the net force on the loop is 2.04x10
N
.

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