# The square loop of conducting wire shown measures 1.0 m on a side and is moving at a constant velocity of 1.0 m/s to the left. ?

The square loop of conducting wire shown measures 1.0 m on a side and is moving at a constant velocity of 1.0 m/s to the left. At t=0, the left edge of the square just begins to enter the uniform magnetic field, which has a field strength of 100 T. The field is 1.5 meters wide.

a. What is the magnitude of the emf induced in the wire loops at t=.05 sec?

b. What is the emf induced in the wire loop at t= 1.0 sec?

c. If the loop has a resistance of 200 ohms, determine the magnitude and direction of the current in the loop at t= 0.5 sec.

d. Determine the magnitude and direction of current in the loop at t= 2.0 sec.

### 1 Answer

- FiremanLv 72 months ago
a) As v = 1 m/s

=>∆A/∆t = 1 m^2/sec

=>∆A/∆t (for t = 0.05) = 1 x 0.05 = 0.05 m^2/sec

BY V = -N∆BA/∆t {-ve just indicating the direction}

=>V = 1 x 100 x 0.05 {as N= 1}

=>V = 5 Voltb) For t = 1 sec

=>∆A/∆t = 1 m^2/sec

BY V = -N∆BA/∆t {-ve just indicating the direction}

=>V = 1 x 100 x 1 {as N= 1}

=>V = 100 Volt

c) For t = 0.5 sec

=>∆A/∆t = 0.5 m^2/sec

BY V = -N∆BA/∆t {-ve just indicating the direction}

=>V = 1 x 100 x 0.5 {as N= 1}

=>V = 50 Volt

Thus By V = iR

=>i = V/R

=>i = 50/200

=>i = 0.25 amp anticlockwise {By right hand rule}

d) At t =2 sec

=>The condition will be same as that of t = 0.5 sec, since the width of the field is 1.5m and the loop have traveled 2 meter.

For t = 2 sec

=>∆A/∆t = 0.5 m^2/sec

BY V = -N∆BA/∆t {-ve just indicating the direction}

=>V = 1 x 100 x 0.5 {as N= 1}

=>V = 50 Volt

Thus By V = iR

=>i = V/R

=>i = 50/200

=>i = 0.25 amp anticlockwise {By right hand rule}

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