Chapter 27, Problem 80A

To determine

To Find: Current in wire which is placed perpendicular to the earth’s magnetic field.

Answer to Problem 80A

  $30.05\text{A}$

Explanation of Solution

Given data:

Magnetic field due to Earth, $B=30.5\times {10}^{-6}\text{T}$

Length of the wire, $l\text{ =1}\text{.2 m}$

Force on the wire, $F=\text{1}\text{.1}\times {\text{10}}^{-3}\text{N}$

Angle, $\theta \text{=}\text{90}°$

Formula used:

When a current carrying wire is placed in a magnetic field, it experiences a force. The current flowing in the wire can be calculated using the formula of that force.

  $F=IlB\sin \theta$

Here, I is the current.

Calculation:

Substitute the values and solve:

  $\begin{array}{l}F=IlB\sin \theta \\ 1.1\times {10}^{-3}\text{N= }I\times \text{1}\text{.2 m}\times 30.5\times {10}^{-6}\text{T}\times \text{sin 90}°\\ I=30.05\text{A}\end{array}$

Conclusion:

The current in wire is $30.05\text{A}$ .