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The plane of a 5.0 cm * 8.0 cm rectangular loop of wire is parallel to a 0.19-T magnetic field. The loop carries a current of 6.2 A. (a) What torque acts on the loop? (b) What is the magnetic moment of the loop? (c) What is the maximum torque that can be obtained with the same total length of wire carrying the same current in this magnetic field?

Short Answer

Expert verified

(a)torque on the loop isτ=4.7×103Nm

(b) the magnetic moment of the loop isμ=0.025Am2

(c) maximum torque that can be obtained isτ=6.34×103N/m

Step by step solution

01

Step 1:

A=5.0cm

B=8.0cm

B=0.19T

I=6.2A

02

Step 2:

(a)for torque on the loop

Force acting is in two parallel wires in loop.one force in and another out but bath has same magnitude,

τ=FrF=laB (1)

Put in equation (1),

τ=ablB

ab=A

role="math" localid="1668234520623" τ=AlB

Put all the known values,

T=(6.0A)(0.050m)(0.080m)(0.19T)=4.7×103NmT=4.7×103Nm

Thus, torque on the loop isτ=4.7×103Nm

03

Step (3)

(b)for the magnetic moment of the loop

Magnetic moment is,

μ=IAμ=(6.2A)(0.050m)(0.080m)=0.025Am2μ=0.025Am2

Thus, the magnetic moment of the loop is

04

Step 4:

(c)for maximum torque,

the maximum area when loop is circular,

2ab + 2b = 2ττR

R is radius,

R=a+bττ

Area of the loop,

A=πR2=πa+bπ2=(0.13m)2π=5.38×103m2

And the torque is,

T=(6.2A)5.38×103m2(0.19T)=6.34×103N/mT=6.34×103N/m

Thus, maximum torque that can be obtained isτ=6.34×103N/m

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