A wire with linear density (mass per unit length) of 1 gram/cm is placed on a ho

Question

A wire with linear density (mass per unit length) of 1 gram/cm is placed on a horizontal surface with a coefficient of kinetic friction of 0.20. The wire carries a current of 1.50 A eastward. In a vertical magnetic field, the wire moves horizontally to the north with a constant speed of 4.2 m/s. What is the magnitude of the vertical magnetic field that enables the wire to move in this fashion? If the coefficient of kinetic friction becomes 0.80 and current becomes 3.0 A, what is the magnitude of the magnetic force acting on the electrons?

Explanation / Answer

(A) and

Linear density d=1g/cm=0.1kg/m

Co efficient of friction u=0.20

Current i=1.5 A

Velocity of electrons moving v=4.2 m/s

First we find normal force using speed of transverse waves

Speed of transverse waves V=N/d

(4.2)^2=N/0.1

Normal force N=1.764 N

Basing law of motion => F=(co efficient of friction) ×N........(1)

Basing on moving charges and magnetism

Force =Bil.................(2)

Now we compare two equations we get

=> co.efficient of friction ×N=Bil...................(3)

0.20×1.764=B×1.5×1

Therefore the magnitude of magnetic field =0.24T

(B) ans

Co.efficient of friction =0.8

Current i=3 A

We use eq3 we get the magnetic field

(Co efficient of friction) ×N=Bil

0.8×1.764=B×3×1

Therefore the magnitude of magnetic field=0.4704 T

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