First slide

Magnetic field due to current element

Question

A charge particle A of charge 2C has velocity 100 m/s. When it passes through point A and has velocity in the direction shown. The strength of magnetic field at point B due to this moving charge is (r = 2m)

Moderate

A

2.5µT
B

5.0µT
C

2.0µT
D

3.7µT

Solution

$\vec{B} = \frac{μ_{o}}{4 π} q \frac{\vec{V} X \vec{r}}{r^{3}} = \frac{μ_{o}}{4 π} \frac{q V \sin θ}{r^{2}} = \frac{10^{- 7} \times 2 \times 100 \times 0.5}{{(2)}^{2}} = 2.5 x 10^{- 6} T$

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