First slide

Magnetic field due to current element

Question

From a cylinder of radius R, a cylinder of radius R/2 is removed, as shown in figure. Current flowing in the remaining cylinder is I. Then, magnetic field strength is

Difficult

A

zero at point A
B

zero at point B
C

$\frac{μ_{0} I}{2 π R}$ at point A
D

$\frac{μ_{0} I}{3 π R}$ at point B

Solution

Current density: $ρ = \frac{I}{π R^{2} - π {(R / 2)}^{2}} \Rightarrow ρ = \frac{4 I}{3 π R^{2}}$

Current in smaller cylinder (if there were): $I_{1} = ρ π {(\frac{R}{2})}^{2} = \frac{I}{3}$
Current in smaller cylinder (if there were) : $I_{2} = I + \frac{I}{3} = \frac{4 I}{3}$
For $A : B_{A} = B_{w h o l e - c y l i n d e r} - B_{s m a l l - c y l i n d e r}$
$\Rightarrow B_{A} = 0 - \frac{μ_{0} (I / 3)}{2 π (R / 2)} = \frac{- μ_{0} I}{3 π R}$
For $B : B_{B} = B_{w h o l e - c y l i n d e r} - B_{s m a l l - c y l i n d e r}$
$= \frac{μ_{0} (\frac{4 I}{3}) (R / 2)}{2 π R^{2}} - 0 = \frac{μ_{0} I}{3 π R}$ .

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