Question #179987

Consider the two dipoles that are located on a line with dipole moments - 5 × 10−9 𝐶. 𝑚 and 9 × 10−9 𝐶. 𝑚, Find the Electric potential at the center of the line. Now, replace the electric dipoles with the magnetic dipoles such that their dipole moments are 5 × 10−9 𝐴. 𝑚2 and 9 × 10−9 𝐴. 𝑚2 . Find the magnetic field at the center. 


1
Expert's answer
2021-04-19T07:21:48-0400

Given,

p1=5×109𝐶.𝑚p_1=- 5\times10^{−9} 𝐶. 𝑚

p2=9×109𝐶.𝑚p_2=9\times10^{−9} 𝐶. 𝑚

Electric potential at the center of the line

V=p4πϵo(r2a2)V=\frac{p}{4\pi \epsilon_o (r^2-a^2)}

Hence, net potential at the center (V)=V1+V2(V)=V_1+V_2


=5×1094πϵo(r2a2)+9×1094πϵo(r2a2)=\frac{-5\times 10^{-9}}{4\pi \epsilon_o (r^2-a^2)}+\frac{9\times 10^{-9}}{4\pi \epsilon_o (r^2-a^2)}


=4×1094πϵo(r2a2)=\frac{4\times 10^{-9}}{4\pi \epsilon_o (r^2-a^2)}

Magnetic dipole at the center of dipole

B=B1+B2B=B_1+B_2

=μom14π(dl)2+μom24π(dl)2=\frac{\mu_o m_1}{4\pi (d-l)^2}+\frac{\mu_o m_2}{4\pi (d-l)^2}

Now, substituting the values,

=μo5×1094π(dl)2+μo9×1094π(dl)2=\frac{\mu_o 5\times 10^{-9}}{4\pi (d-l)^2}+\frac{\mu_o 9\times 10^{-9}}{4\pi (d-l)^2}


=14×107×109(dl)2=\frac{14\times 10^{-7}\times 10^{-9}}{(d-l)^2}


=14×1016(dl)2=\frac{14\times 10^{-16}}{(d-l)^2}



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