Chapter 30: Q96P (page 902)

A square loop of wire is held in a uniformB=0.24Tmagnetic field directed perpendicular to the plane of the loop. The length of each side of the square is decreasing at a constant rate of 5.0 cm/s. What emf is induced in the loop when the length is 12 cm ?

Short Answer

Expert verified

The induced in the loop is, $|ε| = 2.9 \times 10^{- 3} V$

Step by step solution

Given

$B = 0.24 T \frac{d l}{d t} = 5.0 c m / s = 0.05 m / s l = 12 c m = 0.12 m$

Understanding the concept

By using the equation of Faraday’s law and the equation of magnetic flux related to magnetic field and area of the loop, we can find the emfinduced in the loop.

Formula:

$ε = - \frac{d ϕ_{B}}{d t}$

Calculate the emf induced in the loop

The magnetic flux can be written as

$ϕ_{B} = B A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)$

For the square loop the area is is $A = l^{2}$ . So, the rate of area is

$\frac{d A}{d t} = 2 l \frac{d l}{d t}$

Thus, according to faradays law, theis

$ε = - \frac{d ϕ}{d t}$

From equation (1) theis

$ε = - \frac{d (B A)}{d t}$

$ε = - B \frac{d A}{d t}$

As from above equation

$ε = - B \times 2 l \frac{d l}{d t}$

By substituting the value we get

$ε = - 0.24 \times 2 \times 0.12 \times 0.05$

$ε = - 0.0029$

$|ε| = 2.9 \times 10^{- 3} V$

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A square loop of wire is held in a uniformB=0.24Tmagnetic field directed perpendicular to the plane of the loop. The length of each side of the square is decreasing at a constant rate of 5.0 cm/s. What emf is induced in the loop when the length is 12 cm ?

Short Answer

Step by step solution

Given

Understanding the concept

Calculate the emf induced in the loop

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