Chapter 12, Problem 12.1CYU

Check Your Understanding Using Example 12.1, at what distance would P have to be to measure a magnetic field half of tire given answer?

To determine

The distance at which the magnetic field is half the original magnetic field.

Answer to Problem 12.1CYU

The distance at which the magnetic field is half the original magnetic field is $1.414\text{m}$.

Explanation of Solution

Given:

The value of magnetic field is $2\times {10}^{-9}\text{T}$.

The length of wire is $1\text{cm}$.

The current in the wire is $2\text{A}$.

The value of angle is $89.4°$.

The distance of wire from point is $1\text{m}$.

Formula used:

The expression for new magnetic field at point P is:

  $B_1=\frac{{\mu }_0}{4\pi }\frac{Idl\sin \theta }{r_1^2}$   ...... (1)

Here, $B_1$ is the new magnetic field, $I$ is the current in the wire, $r_1$ is the distance of point P where magnetic field is half, ${\mu }_0$ is the permeability for free space and $dl$ is the path element.

The expression for original magnetic field at point P.

  $B=\frac{{\mu }_0}{4\pi }\frac{Idl\sin \theta }{r^2}$  ......(2)

Here, $B$ is the original magnetic field and $r$ is the initial distance of point P.

New magnetic field is half of the original magnetic field.

  $B_1=\frac{B}{2}$

Here, $B$ is the original magnetic field.

Take ratio of equation (2) and equation (1).

  $\frac{B}{B_1}=\frac{\left(\frac{{\mu }_0}{4\pi }\frac{Idl\sin \theta }{r^2}\right)}{\left(\frac{{\mu }_0}{4\pi }\frac{Idl\sin \theta }{r_1^2}\right)}$  ......(3)

Calculation:

Substitute $\frac{B}{2}$ for $B_1$ and $1\text{m}$ for $r$ in equation (3).

  $\begin{array}{c}\frac{B}{\left(\frac{B}{2}\right)}=\frac{\left(\frac{{\mu }_0}{4\pi }\frac{Idl\sin \theta }{{\left(1\text{m}\right)}^2}\right)}{\left(\frac{{\mu }_0}{4\pi }\frac{Idl\sin \theta }{r_1^2}\right)}\\ 2=\left(\frac{r_1^2}{1{\text{m}}^{\text{2}}}\right)\\ r_1=1.414\text{m}\end{array}$

Conclusion:

Thus, the distance at which the magnetic field is half the original magnetic field is $1.414\text{m}$.