Question

A charge particle of charge q = 2 C has velocity $v=100{ms}^{-1}$. When it passes point A adn has velocity in the direction shown. The strength of magnetic field at the point B due to this moving charge is (r = 2m)

• A. $2.5\mu T$
• B. $5.0\mu T$
• C. $2.0\mu T$
• D. None of these

Answer 1

The correct option is A $2.5\mu T$

Given :charge(q)=2C,v=100m/s, r=2m,$\theta$=30v=100ms−1

Solution: We know the formula of magnetic field on moving charge is :

$B=\frac{{\mu }_{0}M}{4\pi r^3}$

And as we know the know the formula of magnetic moment in terms of q &v

$M=q(\overrightarrow{v}\times \overrightarrow{r})$

$B=\frac{{\mu }_{0}q\left(vrsin{30}^0\right)}{4\pi r^3}=\frac{{10}^{-7}\times 2\times 100\times 1}{4\times 2}=2.5\mu T$

Hence the correct answer is A