Chapter 26, Problem 73A

To determine

To find: Field strength of a current carrying conductor in present of magnetic field.

Answer to Problem 73A

  $\boxed{B=0.016{\text{Wb/m}}^{\text{2}}}$ .

Explanation of Solution

Given:

The length of the wire $440\text{cm}$ .

The force on the wire is $0.55\text{N}$ .

Amount of current flow $7.7\text{A}$ .

The angle is $\text{θ}={90}^{\circ }$ .

Formula:

The force experience by a wire having length $l$ in present of magnetic field

  $F=ilB\sin \text{θ}$  ...... (1)

Here, $i$ is the current flow the wire, $F$ is the force, $l$ is the length of the wire and $B$ is the magnetic field strength.

Calculation:

Simplified equation (1)

  $B=\frac{F}{il\sin \theta }$…… (2)

Substitute $440\text{cm}$ for $l$ and $0.55\text{N}$ for $F$ and $7.7\text{A}$ for $\text{i}$ in equation (1)

  $\begin{array}{c}B=\frac{0.55}{7.7\times 440\times \sin 90}\\ =\frac{0.55}{7.7\times 4.40\text{m}}\\ \boxed{B=0.016{\text{Wb/m}}^{\text{2}}}\end{array}$

Conclusion:

Hence, the required magnetic field strength is $\boxed{B=0.016{\text{Wb/m}}^{\text{2}}}$ .