**3. Problem Description:**A square coil of wire (36.0 cm long on each side with 50 loops) moves at a constant speed of 0.40 m/s throughout the problem. The loop of wire moves from a region of zero magnetic field to a region of magnetic field of strength 0.84 T directed into the page as shown below. The square loop has a resistance of 7.50 Ω. The loop is completely out of the magnetic field 2.50 s after first beginning to enter the region.**Diagram:**- At time \( t=0 \), the square coil is outside the dotted region representing the magnetic field.- At time \( t=2.50 \, \text{s} \), the square coil has exited the magnetic field region.- The velocity \( v \) is indicated, moving from left to right.**Given:**- \( N = 50 \) (number of loops)- \( v = 0.40 \, \text{m/s} \) (velocity)- \( B = 0.84 \, \text{T} \) (magnetic field strength)- \( R = 7.5 \, \Omega \) (resistance)- \( L = 0.36 \, \text{m} \) (length of side)- \( t = 2.5 \, \text{s} \) (time)**Equations:**- \( \mathcal{E} = - \frac{d\Phi}{dt} \) (Faraday’s Law of Induction)- \( I = \frac{\mathcal{E}}{R} \) (Ohm’s Law)**Tasks:**a) From its position at \( t=0 \), how much time does it take for the square coil to get completely into the region of magnetic field?b) What is the magnetic flux through the coil when it has fully entered the region with magnetic field?c) Find the magnitude of the current in the square coil: i) As the coil enters the magnetic field region ii) As it moves through the region iii) As the coil exits the region.d) Find the direction of the current in the loop during these three time intervals (in part c).

Given:length of 1 side of the coil, L = 0.36 mtotal loops, n = 50velocity of the coil, v = 0.40…

ne so = 0.40w/s Razisn 3. A square coil of wire (36.0 cm long on each side with 50 loops) moves at a constant L = 0.3m B=0.841 € = 2.5s speed of 0.40 m/s throughout the problem. The loop of wire moves from a region of zero magnetic field to a region of magnetic field of strength 0.84 T directed into the page as shown below. The square loop has a resistance of 7.50 92, The loop is completely out of the magnetic field 2.50 s after first beginning to enter the region. t=0 s

ne so = 0.40w/s Razisn 3. A square coil of wire (36.0 cm long on each side with 50 loops) moves at a constant L = 0.3m B=0.841 € = 2.5s speed of 0.40 m/s throughout the problem. The loop of wire moves from a region of zero magnetic field to a region of magnetic field of strength 0.84 T directed into the page as shown below. The square loop has a resistance of 7.50 92, The loop is completely out of the magnetic field 2.50 s after first beginning to enter the region. t=0 s

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Concept explainers

Magnetic Field Of Coaxial Cable

The word coaxial means same axis. So for a long straight cable to be coaxial, it must have concentric cylindrical shaped conductor around it. Coaxial cable (popularly called ‘coax’) has various applications ranging from current conductors to radiofrequency signal transmitters. This cable has lower emission losses and provides protection from electromagnetic interference which allows signals with less power to transmit over longer distances.For example, currents produced in the pickup of a stereo turntable generally travel to the amplifier through a coaxial cable.The self-inductance of a coaxial cable is another important characteristic, because it influences the propagation of electrical signals in the cable.

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**3. Problem Description:**

A square coil of wire (36.0 cm long on each side with 50 loops) moves at a constant speed of 0.40 m/s throughout the problem. The loop of wire moves from a region of zero magnetic field to a region of magnetic field of strength 0.84 T directed into the page as shown below. The square loop has a resistance of 7.50 Ω. The loop is completely out of the magnetic field 2.50 s after first beginning to enter the region.

**Diagram:**

- At time \( t=0 \), the square coil is outside the dotted region representing the magnetic field.
- At time \( t=2.50 \, \text{s} \), the square coil has exited the magnetic field region.
- The velocity \( v \) is indicated, moving from left to right.

**Given:**
- \( N = 50 \) (number of loops)
- \( v = 0.40 \, \text{m/s} \) (velocity)
- \( B = 0.84 \, \text{T} \) (magnetic field strength)
- \( R = 7.5 \, \Omega \) (resistance)
- \( L = 0.36 \, \text{m} \) (length of side)
- \( t = 2.5 \, \text{s} \) (time)

**Equations:**
- \( \mathcal{E} = - \frac{d\Phi}{dt} \) (Faraday’s Law of Induction)
- \( I = \frac{\mathcal{E}}{R} \) (Ohm’s Law)

**Tasks:**

a) From its position at \( t=0 \), how much time does it take for the square coil to get completely into the region of magnetic field?

b) What is the magnetic flux through the coil when it has fully entered the region with magnetic field?

c) Find the magnitude of the current in the square coil:
i) As the coil enters the magnetic field region
ii) As it moves through the region
iii) As the coil exits the region.

d) Find the direction of the current in the loop during these three time intervals (in part c).

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