Practicing Success
A steady current I flow through a long straight wire of radius 'a'. The current is uniformly distributed across its cross section. The ratio of the magnetic fields due to the wire at distance $\frac{a}{4}$ and 3a respectively from the axis of the wire is |
3 : 4 4 : 3 2 : 3 1 : 4 |
3 : 4 |
Magnetic field inside the wire at a distance r is $ B = \frac{\mu_0 Ir}{2\pi a^2}$ $ r =\frac{a}{4} , B_1 = \frac{\mu_0 I}{8\pi a}$ Magnetic field outside wire at a distance r is $ B = \frac{\mu_0 I}{2\pi r}$ $ r=3a , B_2 = \frac{\mu_0 I}{6\pi a}$ $ \frac{B_1}{B_2} = \frac{3}{4}$
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