Question

What is the magnetic flux through the inner small coil when current \[I_2\] flows in the outer large coil?

पर्याय

• \[\Phi_1=\pi r_2^2B_2=\frac{\mu_0\pi r_2I_2}{2}\]

• \[\Phi_1=\pi r_1^2B_2=\frac{\mu_0\pi r_1^2I_2}{2r_2}\]

• \[\Phi_1=\frac{B_2}{\pi r_1^2}\]

• \[\Phi_1=\frac{\mu_0\pi r_1^2}{2r_2I_2}\]

Answer 1

The uniform-field approximation gives flux as \[\Phi_1=\pi r_1^2B_2\] . Substituting the outer coil’s central field yields \[\Phi_1=\frac{\mu_0\pi r_1^2I_2}{2r_2}\]