The time dependent magnetic dipole moment is driven by the rotation of the star, so it is natural that rotation would provide the energy that goes into radiation. (You can check this by computing the torque.) Indeed, the energy in the magnetic field is much too small to power the emission.

The B-field, on the other hand, cannot just disappear (magnetic field lines in ideal MHD cannot just disappear). The B field decays by ohmic diffusion $$ \frac{\partial B}{\partial t} = \frac{c^2}{4\pi\sigma}\nabla^2 B $$ This gives a deacy time $$ \tau = \frac{4\pi\sigma}{c^2}\frac{R^2}{\pi^2} $$ Using $R=10$ km and $\sigma=6\cdot 10^{22}$ $s^{-1}$ (this is the conductivity cgs units) G. Baym, C. Pethick, and D. Pines, Nature, 224, 673, (1969) get $\tau=4\cdot 10^6$ yr, several million years.

Postcript: A useful review is Petri, https://arxiv.org/abs/1608.04895v1 . Among many other things the author provides estimates of the energies involved. For a mili-second pulsar the gravitational energy is $2.6 \cdot 10^{46}$ J, the rotational energy is $3.2 \cdot 10^{45}$ J, the magnetic energy is $1.6 \cdot 10^{28}$ J, and the thermal energy is $3.4 \cdot 10^{40}$ J.

The time dependent magnetic dipole moment is driven by the rotation of the star, so it is natural that rotation would provide the energy that goes into radiation. (You can check this by computing the torque.) Indeed, the energy in the magnetic field is much too small to power the emission.

The B-field, on the other hand, cannot just disappear (magnetic field lines in ideal MHD cannot just disappear). The B field decays by ohmic diffusion $$ \frac{\partial B}{\partial t} = \frac{c^2}{4\pi\sigma}\nabla^2 B $$ This gives a decay time $$ \tau = \frac{4\pi\sigma}{c^2}\frac{R^2}{\pi^2} $$ Using $R=10$ km and $\sigma=6\cdot 10^{22}$ $s^{-1}$ (this is the conductivity cgs units) G. Baym, C. Pethick, and D. Pines, Nature, 224, 673, (1969) get $\tau=4\cdot 10^6$ yr, several million years.

Postscript: A useful review is Petri, https://arxiv.org/abs/1608.04895v1 . Among many other things the author provides estimates of the energies involved. For a mili-second pulsar the gravitational energy is $2.6 \cdot 10^{46}$ J, the rotational energy is $3.2 \cdot 10^{45}$ J, the magnetic energy is $1.6 \cdot 10^{28}$ J, and the thermal energy is $3.4 \cdot 10^{40}$ J.

The time dependent magnetic dipole moment is driven by the rotation of the star, so it is natural that rotation would provide the energy that goes into radiation. (You can check this by computing the torque.) Indeed, the energy in the magnetic field is much too small to power the emission.

The B-field, on the other hand, cannot just disappear (magnetic field lines in ideal MHD cannot just disappear). The B field decays by ohmic diffusion $$ \frac{\partial B}{\partial t} = \frac{c^2}{4\pi\sigma}\nabla^2 B $$ and in a neutron star the time scale for this is millions of years, see G. Baym, C. Pethick, and D. Pines, Nature, 224, 673, (1969).

The time dependent magnetic dipole moment is driven by the rotation of the star, so it is natural that rotation would provide the energy that goes into radiation. (You can check this by computing the torque.) Indeed, the energy in the magnetic field is much too small to power the emission.

The B-field, on the other hand, cannot just disappear (magnetic field lines in ideal MHD cannot just disappear). The B field decays by ohmic diffusion $$ \frac{\partial B}{\partial t} = \frac{c^2}{4\pi\sigma}\nabla^2 B $$ This gives a deacy time $$ \tau = \frac{4\pi\sigma}{c^2}\frac{R^2}{\pi^2} $$ Using $R=10$ km and $\sigma=6\cdot 10^{22}$ $s^{-1}$ (this is the conductivity cgs units) G. Baym, C. Pethick, and D. Pines, Nature, 224, 673, (1969) get $\tau=4\cdot 10^6$ yr, several million years.

Postcript: A useful review is Petri, https://arxiv.org/abs/1608.04895v1 . Among many other things the author provides estimates of the energies involved. For a mili-second pulsar the gravitational energy is $2.6 \cdot 10^{46}$ J, the rotational energy is $3.2 \cdot 10^{45}$ J, the magnetic energy is $1.6 \cdot 10^{28}$ J, and the thermal energy is $3.4 \cdot 10^{40}$ J.