Chapter 21, Problem 7P

A 16-cm-diameter circular loop of wire is placed in a 0.50-T magnetic field. (a) When the plane of the loop is perpendicular to the field lines, what is the magnetic flux through the loop? (b) The plane of the loop is rotated until it makes a 42° angle with the field lines. What is the angle θin Eq. 21-1for this situation? (c) What is the magnetic flux through the loop at this angle?

To determine

(a) To determine:

Magnetic flux through the loop

Answer to Problem 7P

Solution:

Magnetic flux through the loop is $1\times {10}^{-2}\text{Wb}$

Explanation of Solution

Given:

Diameter of the loop, $d=16\text{cm}$

Magnetic field strength, $B=0.50\text{T}$

Formula used:

The magnetic field through the loop is given as

$\varphi =BA\cos \theta$

Here,

$B$ is magnetic field strength

$A$ is area

$\theta$ is the angle

Calculation:

The magnetic field through the loop is given as

$\varphi =BA\cos \theta$

Plugging the values in the above equation

$\begin{array}{c}\varphi =0.50\times \frac{\pi }{4}\left({0.16}^2\right)\cos 0\\ =1\times {10}^{-2}\text{Wb}\end{array}$

To determine

(b) To determine:

Angle $\theta$ for this situation

Answer to Problem 7P

Solution:

Angle $\theta$ for this situation is $48°$

Explanation of Solution

Given:

The plane of the loop is rotated until it makes $42°$ with the field lines

Calculation:

The angle of the loop is calculated as

$\begin{array}{c}\theta =90-42°\\ =48°\end{array}$

To determine

(c) To determine:

Magnetic flux through the loop

Answer to Problem 7P

Solution:

Magnetic flux through the loop is $6.7\times {10}^{-3}\text{Wb}$

Explanation of Solution

Given:

Diameter of the loop, $d=16\text{cm}$

Magnetic field strength, $B=0.50\text{T}$

Angle, $\theta =48°$

Formula used:

The magnetic field through the loop is given as

$\varphi =BA\cos \theta$

Here,

$B$ is magnetic field strength

$A$ is area

$\theta$ is the angle

Calculation:

The magnetic field through the loop is given as

$\varphi =BA\cos \theta$

Plugging the values in the above equation

$\begin{array}{c}\varphi =0.50\times \frac{\pi }{4}\left({0.16}^2\right)\cos 48°\\ =6.7\times {10}^{-3}\text{Wb}\end{array}$