Problem

A wire carrying a current of $10 \, \mathrm{mA}$ experiences a force of $2 \, \mathrm{N}$ in a uniform magnetic field. What will be the force on the wire if the current rises to $30 \, \mathrm{mA}$?

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Data

• Initial force: $F_i = 2 \, \mathrm{N}$,
• Initial current: $I_i = 10 \, \mathrm{mA} = 10 \times 10^{-3} \, \mathrm{A}$,
• Final current: $I_f = 30 \, \mathrm{mA} = 30 \times 10^{-3} \, \mathrm{A}$.

To find:

• Final force: $F_f = ?$.

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Prerequisite Concepts

The magnetic force on a current-carrying conductor is directly proportional to the current. The force is given by:

$$F = B I L \sin \theta$$

where $B$ is the magnetic field strength, $I$ is the current, $L$ is the length of the conductor, and $\theta$ is the angle between the conductor and the magnetic field.

Dividing the final and initial force expressions:

$$\frac{F_f}{F_i} = \frac{I_f}{I_i}$$

Rearranging for $F_f$:

$$F_f = F_i \cdot \frac{I_f}{I_i}$$

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Solution

1. Substitute the known values into the formula:

$$F_f = 2 \, \mathrm{N} \cdot \frac{30 \times 10^{-3}}{10 \times 10^{-3}}$$

2. Simplify:

$$F_f = 2 \cdot \frac{30}{10} = 2 \cdot 3 = 6 \, \mathrm{N}.$$

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Answer

The force on the wire when the current rises to $30 \, \mathrm{mA}$ is:

$$F_f = 6 \, \mathrm{N}.$$