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Chapter 20: Q73P (page 863)

A bar magnet whose magnetic dipole moment is 15鈥碱.尘2is aligned with an applied magnetic field of 4鈥涂. How much work must you do to rotate the bar magnet 180掳to point in the direction opposite to the magnetic field?

Short Answer

Expert verified

The work done to rotate the bar magnet 180掳to point in the direction opposite to the magnetic field is 120 J.

Step by step solution

01

Given Information

The dipole moment of the magnet is 渭=15鈥碱.尘2, and the magnetic field is B=4鈥涂. T. The bar rotates at an angle 胃=180掳.

02

Work done

The directions of the bar rotation and the magnetic torque are the same, therefore, the work done is non-zero and equals to the difference between the two potential energies at the angle 胃1=0and 胃2=180掳given as:

W=螖U2鈭�螖U1鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌�鈰�1

Where the potential energy for a magnetic dipole is given by

U=鈭�渭鈫�鈰�B鈫�=鈭�渭Bcos胃

Therefore, the equation can be written as:

W=螖U1鈭�螖U2=鈭�渭Bcos胃2鈭�鈭�渭Bcos胃1=渭Bcos胃1鈭�cos胃2鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆��鈰�2

03

Calculate the work done

Substitute the given values of 渭,B,胃1and 胃2into equation (2) to get the work done.

W=渭Bcos胃1鈭�cos胃2=15鈰�4cos0掳鈭�cos180掳=601鈭�鈭�1=120鈥塉

Thus, the work done is 120 J.

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Most popular questions from this chapter


In Figure 20.128 on the left is a region of uniform magnetic field B1into the page, and adjacent on the right is a region of uniform magnetic field B2 also into the page. The magnetic field B2is smaller than B1(B2<B1) . You pull a rectangular loop of wire of length w, height h, and resistance R from the first region into the second region, on a frictionless surface. While you do this you apply a constant force F to the right, and you notice that the loop doesn鈥檛 accelerate but moves with a constant speed.

Calculate this constant speed v in terms of the known quantities B1, B2, w, h, R and F , and explain your calculation carefully. Also show the approximate surface-charge distribution on the loop.

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