First slide

Magnetic field due to a current carrying circular ring

Question

A long wire carries a current of 20 A along the axis of solenoid. The field due to the solenoid is 4 mT. The resultant field at a point 3 mm from the solenoid axis is

Easy

A

1.33mT
B

4.2mT
C

5 . 33 mT
D

2 .87 mT

Solution

Field due to the wire_B is given by
$\begin{aligned} B_{wire} = \frac{μ_{0}}{4 π} \times \frac{2 i}{r} = 10^{- 7} \times \frac{2 \times 20}{3 \times 10^{- 3}} T \\ = \frac{4}{3} \times 10^{- 3} T = \frac{4}{3} mT \end{aligned}$
Field due to solenoid is 4 mT
$\begin{array}{ll} ∴ Net magnetic field = \sqrt{[{(B_{W})}^{2} + {(B_{S})}^{2}]} \\ B = {[(4 / 3)^{2} + (4)^{2}]}^{1 / 2} = 4 \cdot 2 mT \end{array}$

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