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×$$10$$−$$7T$$. Then find the value of p.

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×$$10$$−$$7T$$. Then find the value of p.

Infinity Learn · Accepted Answer

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.

Magnetic field due to current element

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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 $p$ .

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Magnetic field due to current element

Remember concepts with our Masterclasses.

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×10−7T. Then find the value of p.

The moving string behaves like a current, i=qt=qL×Lt=λ×v = (40×10−6C/m)(300 m/sec)=1.2×10−2 AThen B=μ0i2πr=2×10−7×1.2×10−20.005 =4.8×10−7tesla.

Ready to Test Your Skills?

free study material

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 $p$ .