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We move a square loop of wire 2.3 cm on a side into a region of uniform magnetic field of 1.5 T with the plane of the loop perpendicular to the direction of the field. The loop is moving with a constant velocity of 2.1 m/s into the field region starting from a point outside the field. If the loop has a resistance of 3 Ω, find the force on the loop as it just enters the magnetic field region. (Enter the magnitude.)

Respuesta :

Answer:

Force will be [tex]F=8.33\times 10^{-4}N[/tex]

Explanation:

We have given side of wire d = 2.3 cm = 0.023 m

Magnetic field B = 1.5 T

Velocity v = 2.1 m/sec

Resistance R = 3 ohm

We know that current is given by [tex]i=\frac{Bvd}{R}=\frac{1.5\times 2.1\times 0.023}{3}=0.02415A[/tex]

Power is given by [tex]P=i^2R=0.02415^2\times 3=0.00174watt[/tex]

We know that power is also given by [tex]P=Fv[/tex]

So [tex]0.00174=F\times 2.1[/tex]

[tex]F=8.33\times 10^{-4}N[/tex]

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