One coulomb of point charge moving with a uniform velocity 10 $\hat{x}$ m/s enters the region x ≥ 0 having a magnetic flux density $\overrightarrow{B}=\left(10y\hat{x}+10x\hat{y}+10\hat{z}\right)T$. The magnitude of force on the charge at x = 0+ is ______ N. ( $\widehat{\mathrm{ y}}$ and $\hat{z}$are unit vectors along x-axis, y-axis, and z-axis, respectively.)

The correct answer is 100.

To find the magnitude of the force acting on the point charge as it enters the magnetic field region, we use the magnetic force (Lorentz force) formula:

$\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$

Where the given parameters are:
Charge, $q=1\text{C}$
Velocity, $\overrightarrow{v}=10\hat{x}\text{m/s}$
Magnetic flux density, $\overrightarrow{B}=(10y\hat{x}+10x\hat{y}+10\hat{z})\text{T}$

Substitute the velocity and magnetic field vectors into the force equation:
$\overrightarrow{F}=1\cdot [10\hat{x}\times (10y\hat{x}+10x\hat{y}+10\hat{z}\left)\right]$

Compute the cross product using the unit vector relationships:
$\hat{x}\times \hat{x}=0$
$\hat{x}\times \hat{y}=\hat{z}$
$\hat{x}\times \hat{z}=-\hat{y}$

Applying these to the calculation:
$\overrightarrow{F}=100x(\hat{x}\times \hat{y})+100(\hat{x}\times \hat{z})$
$\overrightarrow{F}=100x\hat{z}-100\hat{y}\text{N}$

Now, evaluate the force precisely as the charge enters the region at $x=0^{+}$:
$\overrightarrow{F}\left(0^{+}\right)=100\left(0\right)\hat{z}-100\hat{y}=-100\hat{y}\text{N}$

Taking the magnitude of this force:
$\left|\overrightarrow{F}\right|=\sqrt{{(-100)}^2}=100\text{N}$