A wire 1.80 m long carries a current of 13.0 A and makes an angle of 35.0^∘ with a uniform mag

Name: Problem 40, Chapter 28: Fundamentals of Physics
Uploaded: 2019-07-24T21:28:57-08:00
Duration: 1 min 48 s
Channel: Shital Rijal
Description: A wire 1.80 m long carries a current of 13.0 A and makes an angle of 35.0^∘ with a uniform magnetic field of magnitude B= 1.50 T. Calculate the magnetic force on the wire.

step 1 So the question has given us that the length of wire here is 1 .8 meters long and it carries a c

Question

A wire $1.80 \mathrm{~m}$ long carries a current of $13.0 \mathrm{~A}$ and makes an angle of $35.0^{\circ}$ with a uniform magnetic field of magnitude $\underline{B}=$ $1.50 \mathrm{~T}$. Calculate the magnetic force on the wire.

   A wire $1.80 \mathrm{~m}$ long carries a current of $13.0 \mathrm{~A}$ and makes an angle of $35.0^{\circ}$ with a uniform magnetic field of magnitude $\underline{B}=$ $1.50 \mathrm{~T}$. Calculate the magnetic force on the wire.

Fundamentals of Physics

David Halliday,… 10th Edition

Chapter 28, Problem 40

Instant Answer

Solved by Expert Shital Rijal

07/24/2019

Step 1

First, we need to find the component of the magnetic field that is perpendicular to the current in the wire. We can do this using the formula: $B_\perp = B \sin(\theta)$, where $B_\perp$ is the perpendicular component of the magnetic field, $B$ is the magnitude of Show more…

Show all steps

Thanks for your feedback!

A wire $1.80 \mathrm{~m}$ long carries a current of $13.0 \mathrm{~A}$ and makes an angle of $35.0^{\circ}$ with a uniform magnetic field of magnitude $\underline{B}=$ $1.50 \mathrm{~T}$. Calculate the magnetic force on the wire.

Play audio

Feedback

Shital Rijal and 63 other subject Physics 102 Electricity and Magnetism educators are ready to help you.

Ask a new question

Labs

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

Key Concepts

Magnetic Force on a Current-Carrying Wire

This concept involves the calculation of the force experienced by a wire carrying an electric current when placed in a magnetic field. The force is given by the vector cross product F = I L × B, indicating that its magnitude depends on the current, the length of the wire within the field, the strength of the magnetic field, and the sine of the angle between the wire's direction and the magnetic field direction.

Cross Product and Sine Dependence

In the context of magnetic forces, the cross product between the length vector of the wire and the magnetic field vector emphasizes that only the component of the magnetic field that is perpendicular to the current contributes to the force. This results in the magnitude being proportional to the sine of the angle between the wire and the magnetic field, meaning the force reaches its maximum when the wire is perpendicular to the field and becomes zero when the wire is parallel to it.

Please give Ace some feedback