a conducting rod whose length is 25 cm is placed on a u-shaped metal wire that has a resistance of 8.0 ω . the wire and the rod are placed in the plane of the paper. a constant magnetic field of strength 0.40 t is applied out of the paper. an applied force moves the rod to the left with a constant speed of 6.0 m/s. what is the magnitude and direction of the induced current in the wire?

Respuesta :

Answer:

The current is [tex]I = 1 A[/tex]

The direction is anti-clockwise

Explanation:

The diagram for this question is shown on the first uploaded image

From the question we are told that

     the length of the conducting rod is [tex]L = 25 \ cm = \frac{25}{100} = 2.5 \ m[/tex]

     The resistance is  [tex]R = 8 \Omega[/tex]

      The magnetic field is  [tex]B = 0.40\ T[/tex]

        The speed of the rod is [tex]v = 6.0 m/s[/tex]

The emf induced is

          [tex]\epsilon = BLv[/tex]

 substituting values we have

           [tex]\epsilon = 0.40 * 2.5 * 8[/tex]

           [tex]\epsilon = 8V[/tex]

From ohm law the induced current would be

      [tex]I = \frac{\epsilon}{R }[/tex]

 substituting values we have

       [tex]I = \frac{8}{8 }[/tex]      

       [tex]I = 1 A[/tex]

The direction anticlockwise this because according to lenze law the current due to change in magnetic field will act in the opposite direction of the  force causing the magnetic field to change

Ver imagen okpalawalter8

The magnitude of the current will be:

"1 A (Anticlock wise direction)".

Direction and Magnitude

According to the question,

Conducting rod's length, L = 25 cm or,

                                          = 2.5 m

Resistance, R = 8 Ω

Magnetic field, B = 0.40 T

Rod's speed, v = 6.0 m/s

We know the emf induced:

→ [tex]\epsilon[/tex] = BLv

By substituting the values,

      = 0.40 × 2.5 × 8

      = 8 V

By using Ohm's law,

→ I = [tex]\frac{\epsilon}{R}[/tex]

     = [tex]\frac{8}{8}[/tex]

     = 1 A (Anticlock wise)

Thus the approach above is appropriate.

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