   # A coil formed by wrapping 50 turns of wire in the shape of a square is positioned in a magnetic field so that the normal to the plane of the coil makes an angle of 30.0° with the direction of the field. When the magnetic field is increased uniformly from 200 μ T to 600 μ T in 0.400 s, an emf of magnitude 80.0 mV is induced in the coil. What is the total length of the wire in the coil? ### Physics for Scientists and Enginee...

10th Edition
Raymond A. Serway + 1 other
Publisher: Cengage Learning
ISBN: 9781337553278

#### Solutions

Chapter
Section ### Physics for Scientists and Enginee...

10th Edition
Raymond A. Serway + 1 other
Publisher: Cengage Learning
ISBN: 9781337553278
Chapter 30, Problem 7P
Textbook Problem
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## A coil formed by wrapping 50 turns of wire in the shape of a square is positioned in a magnetic field so that the normal to the plane of the coil makes an angle of 30.0° with the direction of the field. When the magnetic field is increased uniformly from 200 μT to 600 μT in 0.400 s, an emf of magnitude 80.0 mV is induced in the coil. What is the total length of the wire in the coil?

To determine
The total length of wire in the coil.

### Explanation of Solution

Given info: Number of turns of wire is 50 , angle of the normal to the direction of field is 30° , Magnetic field is increasing from 200μT to 600μT in 0.400s and emf induced in the coil is 80.0mV .

The rate of increase in magnetic field with respect to time can be given as,

ΔBΔt=B2B1t

Here,

B1 is the initial magnetic field.

B2 is the final magnetic field.

t is the time.

Substitute 200μT for B1 , 600μT for B2 and 0.400s for t in the above equation,

ΔBΔt=(600μT200μT)(1T106μT)0.400s=103T/s

According to Faraday’s law of induction, the emf induced in a coil can be given as,

ε=NΔBΔtAcosθ=NΔBΔta2cosθ

Here,

N is the number of turns in coil

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