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A metal disk of radius 6.0 cm is mounted on a frictionless axle. Current can flow through the axle out along the disk, to a sliding contact on the rim of the disk. A uniform magnetic field B= 1.00 T is parallel to the axle of the disk. When the current is 3.0 A, the disk rotates with constant angular velocity. What's the frictional force at the rim between the stationary electrical contact and the rotating rim?

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lublana

Answer:

0.09 N

Explanation:

We are given that

Radius of disk,r=6 cm=[tex]\frac{6}{100}=0.06 m[/tex]

1 m=100 cm

B=1 T

Current,I=3 A

We have to find the frictional force at the rim between the stationary electrical contact and the rotating rim.

[tex]dF=IBdr[/tex]

[tex]dF=IBdr[/tex]

[tex]\tau=rdF=IBrdr[/tex]

[tex]\tau=\int_{0}^{R}IBr dr[/tex]

[tex]\tau=IB(\frac{R^2}{2}[/tex]

Torque due to friction

[tex]\tau=R\times F[/tex]

Where friction force=F

[tex]R\times F=\frac{IBR^2}{2}[/tex]

[tex]F=\frac{IBR}{2}[/tex]

Substitute the values

[tex]F=\frac{3\times 1\times 0.06}{2}[/tex]

[tex]F=0.09 N[/tex]

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