Chapter 20, Problem 26P

To determine

The strength of magnetic field at a given distance.

To compare: The obtained value with the Earth’s magnetic field.

Answer to Problem 26P

  $B=21.67\times {10}^{-5}\text{T}$

The obtained value of magnetic field is 4.34 times of Earth’s magnetic field.

Explanation of Solution

Given:

Current carried by ajumper cable is 65 A.

Distance is 6 cm.

Formula used:

Magnetic field is,

  $B=\frac{{\mu }_{0}I}{2\pi r}$

Here,

B is magnetic field.

I is current.

r is distance

Calculation:

Here r =6.0cm =0.06m and I =65A

Substituting the values,

  $B=\frac{4\pi \times {10}^{-7}\times 65}{2\pi \times 0.06}$

  $B=21.67\times {10}^{-5}\text{T}$

Earth magnetic field is $5\times {10}^{-5}\text{T}$

So,

  $\begin{array}{l}\frac{B_e}{B}=\frac{5\times {10}^{-5}}{21.67\times {10}^{-5}}=0.23\\ B=4.34B_e\end{array}$

Conclusion:

Therefore, $B=21.67\times {10}^{-5}\text{T}$

The obtained value of magnetic field is 4.34 times of Earth’s magnetic field.