- #1
PhysicsTest
- 238
- 26
- Homework Statement
- To find the magnetic field in a straight current carrying conductor due to sine wave at a distance x on its perpendicular bisector.
- Relevant Equations
- ##B = \frac {\mu_0 I 2a} {4\pi x\sqrt{x^2 + a^2}}##
It is not a direct home work problem, i was thinking if a sine wave current passes through the straight current carrying conductor, what will be the magnetic field. For the DC current I know the formula as below.
##B = \frac {\mu_0 I 2a} {4\pi x\sqrt{x^2 + a^2}}##
Let the current be ##I = I_0\sin(\omega t)## then will the magnetic field be
##B = \frac {\mu_0 I_0\sin(\omega t) 2a} {4\pi x\sqrt{x^2 + a^2}}##
##B = \frac {\mu_0 I 2a} {4\pi x\sqrt{x^2 + a^2}}##
Let the current be ##I = I_0\sin(\omega t)## then will the magnetic field be
##B = \frac {\mu_0 I_0\sin(\omega t) 2a} {4\pi x\sqrt{x^2 + a^2}}##