First slide

Force on a current carrying wire placed in a magnetic field

Question

Magnetic induction at a distance of 20 cm from the axis of long straight wire carrying a current i is 5 µT. Then what is the magnetic induction at a distance 40 cm from the axis of a long straight wire carrying a current 2i?

Moderate

A

3 µT
B

4.5 µT
C

2.5 µT
D

5 µT

Solution

$\begin{array}{l} B = \frac{μ_{o} i}{2 πr} \Rightarrow \frac{B_{2}}{B_{1}} = (\frac{i_{2}}{i_{1}}) (\frac{r_{1}}{r_{2}}) \\ \Rightarrow \frac{B_{2}}{5} = (2) (\frac{20}{40}) \Rightarrow B_{2} = 5 μT \end{array}$

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