# A current carrying wire of length 50 cm is positioned perpendicular to a uniform magnetic field....

## Question:

A current carrying wire of length 50 cm is positioned perpendicular to a uniform magnetic field. I the current is 10.0 A and it is determined that there is a resultant force of 3.0 N on the wire due to the interaction of the current and field, what is the magnetic field strength?

a. 0.6 T

b. 1.50 T

c. 1.85 x 10{eq}^{-3} {/eq} T

d. 6.7 x 10{eq}^{-3} {/eq} T

## Magnetic Force:

When a current carrying wire is kept aligned with the external magnetic field, it doesn't experience any force. However, it the wire is obliquely placed with respect to the external magnetic field, it experiences a magnetic force, which is maximum when the wire is placed perpendicular to the magnetic field.

## Answer and Explanation: 1

**Given Data**

- Length of the wire, {eq}L\ =50\ \text{cm}\ = 0.50\ \text{m}{/eq}

- Angle made by the wire with the magnetic field, {eq}\theta\ = 90^\circ{/eq}

- Current in the wire, {eq}I\ = 10.0\ \text{A} {/eq}

- Magnetic force on the wire, {eq}F\ = 3.0\ \text{N} {/eq}

**Finding the magnetic field strength ( B) in the region**

The magnitude of the force experienced by the wire is expressed as:

- {eq}F\ = I\times L\times B\times \sin \theta{/eq}

- {eq}3.0\ \text{N}\ = 10.0\ \text{A}\times 0.50\ \text{m}\times B\times \sin 90^\circ{/eq}

- {eq}B\ = 0.6\ \text{T} {/eq}

**Correct option is (a)**

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Chapter 18 / Lesson 7Magnetism and electricity are closely related. When current flows in a wire, a magnetic field is generated. In this lesson, we will explore how magnetic fields can exert forces on parallel current-carrying wires.