First slide

Magnetic field due to current element

Question

A long straight non-conducting string carries a charge density of $40 {μC m}^{-1}$ . It is pulled along its length at a speed of 300 m/sec, then the magnetic field at a normal distance of 5 mm
from the moving string is $p \times 10^{- 7} T$ . Then find the value of $p$ .

Moderate

Solution

The moving string behaves like a current,
$i = \frac{q}{t} = \frac{q}{L} \times \frac{L}{t} = λ \times v = (40 \times 10^{- 6} C / m) (300 m / \sec)$ $= 1.2 \times 10^{- 2}$ A
Then $B = \frac{μ_{0} i}{2 π r} = \frac{2 \times 10^{- 7} \times 1.2 \times 10^{- 2}}{0.005}$
$= 4.8 \times 10^{- 7}$ tesla.

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