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AVprozaik [17]
3 years ago
15

Show that the energy of a magnetic dipole m in a magnetic field B is U--m B

Physics
1 answer:
Juli2301 [7.4K]3 years ago
5 0

Answer:

showm

Explanation:

Consider a dipole having magnetic moment 'm' is placed in magnetic field \vec{B} then the torque exerted by the field on the dipole is

\tau = m\times B

\tau=mBsin\alpha

Now to rotate the dipole in the field to its final position the work required to be done is

U=\int \tau d\alpha

U=\int mBsin\alpha d\alpha

U= -mBcos\alpha

U=-\vec{m\times \vec{B}}

Minimum energy mB is for the case when m is anti parallel to B.

Minimum energy -mB is for the case when m is parallel to B.

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Let's solve the problem in the three different steps

1) Uniformly accelerated motion, with acceleration a_1 = 2.65~m/s^2 and for a total time of t_1=17~s. The body is initially at rest, so the distance covered is given by
S= \frac{1}{2}a_1t_1^2=382.9~m
Calling v_f and v_i the final and initial velocity, and since the v_i=0~m/s because the body starts from rest, we can use
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2) Uniform motion. The velocity is constant and it is equal to the final velocity of the first leg: v_2=45~m/s. This is also the average velocity of the second leg. 
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3) Uniformly decelerated motion, with constant deceleration of a_3=-9.39~m/s^2 and for a total time of t_3=4.8~s. Here, the initial velocity of the body is the final velocity of the previous leg, i.e. v_i=45~m/s. Therefore, the distance covered in this leg is given by
S_3=v_i t_3 + \frac{1}{2} a_3 t^2 =107.8~m
The final velocity in this leg is given by
v_f=v_i+at=45~m/s-9.39~m/s^2 \cdot 4.8~s = -0.07~m/s
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S=S_1+S_2+S_3=382.9~m+4320~m+107.8~m=4810.7~m
To find the average velocity, we must "weigh" the average velocity of each leg for the correspondent time of that leg:
v_{ave}= \frac{v_1t_1+v_2t_2+v_3t_3}{t_1+t_2+t_3}=40.8~m/s
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