Explanation Given Info: The mass ( m ) of the uniform rod is 0.0840 kg . The length ( L ) of the uniform rod is 18.0 cm . The uniform magnetic field ( B ) is 0.120 T . The direction of the magnetic field is into the page. The rod is held at the angle ( θ ) of 53.0 ° . The current ( I ) through the rod is 12.0 A . The rod is in rotational equilibrium, therefore the torque due to gravity and the torque to magnetic field must be balanced. Formula to calculate the torque due to the gravitational force is, τ g = 1 2 m g L cos ( θ ) (I) m is the mass of the uniform rod, g is the acceleration due to gravity, L is the length of the uniform rod, and θ is the angle at which the rod is inclined above the horizontal surface. Formula to calculate the torque due to the magnetic field is, τ m = 1 2 I B L 2 (II) I is the current through the uniform rod, and B is the uniform magnetic field in which the rod is place. Formula to calculate the tension of the string is, τ T = T L sin ( θ ) (III) T is the tension of the string. For balancing condition, τ T = τ g + τ m (IV) Substitute equation (I), equation (II), and equation (III) in equation (IV). T L sin ( θ ) = 1 2 m g L cos ( θ ) + 1 2 I B L 2 T L sin ( θ ) = 1 2 L ( m g cos ( θ ) + I B L ) T sin ( θ ) = 1 2 ( m g cos ( θ ) + I B L ) Rearranging the equation, T = ( m g cos ( θ ) + I B L ) 2 sin ( θ ) Substitute 0.0840 kg for m , 9

14th Edition

ISBN: 9780133969290

Author: Hugh D. Young, Roger A. Freedman

Publisher: PEARSON

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Magnetic Force

A magnetic force arises when a charged particle or a current-carrying conductor is placed in a magnetic field. It is created because of the movement of the charged particles.

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Chapter 27, Problem 27.74P

To determine

The tension in the string.

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Chapter 27, Problem 27.73P

Chapter 27, Problem 27.75P

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The tension in the string.

Solution Summary: The author explains how the tension on the string is 0.472n, and the direction of the magnetic field is into the page.

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The tension in the string.

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Chapter 27 Solutions