A long, straight wire of diameter 2.0mm carries a uniformly distributed current of 10A. At what distance from the axis of the wire will the magnitude of B vector be maximum.justify your answer
For 0<r<R=2(mm)0<r<R=2(mm)0<r<R=2(mm)
∫Bdl=μ0I\int Bdl=\mu_0I∫Bdl=μ0I
B⋅2πr=μ0⋅IπR2⋅πr2→B=μ0I2πR2rB\cdot2\pi r=\mu_0\cdot \frac{I}{\pi R^2}\cdot\pi r^2\to B=\frac{\mu_0 I}{2\pi R^2}rB⋅2πr=μ0⋅πR2I⋅πr2→B=2πR2μ0Ir
For r>Rr>Rr>R
B⋅2πr=μ0I→B=μ0I2πrB\cdot2\pi r=\mu_0 I\to B=\frac{\mu_0 I}{2\pi r}B⋅2πr=μ0I→B=2πrμ0I
So, B=Bmax=μ0I2πRB=B_{max}=\frac{\mu_0I}{2\pi R}B=Bmax=2πRμ0I if r=R=2(mm)r=R=2(mm)r=R=2(mm) . Answer
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