# A square current loop 5.00 cm on each side carries a 450 mA current. The loop is in a 0.900 T...

## Question:

A square current loop {eq}5.00\ cm {/eq} on each side carries a {eq}450\ mA {/eq} current. The loop is in a {eq}0.900\ T {/eq} uniform magnetic field. The axis of the loop, perpendicular to the plane of the loop, is {eq}30 ^\circ {/eq} away from the field direction. What is the magnitude of the torque on the current loop?

## Magnetic Torque:

Due to the perpendicular nature of the magnetic force (perpendicular to the axis of an object and the magnetic field), there will be a perpendicular motion that will occur due to this force. This can be interpreted as rotation of the loop about the axis. And since force of rotation is the definition of torque, we can determine the magnetic torque by using the equation:

{eq}\displaystyle \tau = NIAB sin\theta {/eq}

where:

*I*is the current in the loop*A*is the cross-sectional area of the loop- N is the number of turns of the loop
- B is the magnetic field strength
- {eq}\displaystyle \theta {/eq} is the angle between the normal vector of the loop and the field direction.

## Answer and Explanation:

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Given:

- {eq}\displaystyle d = 5\ cm = 0.05\ m {/eq} is the side length of the square

- {eq}\displaystyle I = 450\ mA = 0.45\ A {/eq} is the current...

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Chapter 3 / Lesson 13After watching this video, you will be able to explain what torque is and use an equation to calculate torque in simple situations. A short quiz will follow.