$\text{An electron (mass 9 × 10⁻³¹ kg and charge 1.6 × 10⁻¹⁹ C) moving with speed c/100 (c = speed of light) is injected into a magnetic field B of magnitude 9 × 10⁻⁴ T perpendicular to its direction of motion.}$$\text{We wish to apply a uniform electric field E together with the magnetic field so that the electron does not deflect from its path.}$$\text{Then (speed of light c = 3 × 10⁸ m s⁻¹)}$

The correct option is:
E is perpendicular to B and its magnitude is 27 × 10² V m⁻¹

Step-by-Step Derivation and Logical Reasoning:

1. Condition for No Deflection:
When a charged particle (like an electron) passes through both an electric field $E$ and a magnetic field $B$ without undergoing any deflection, the net Lorentz force acting on it must be zero.
The total Lorentz force is given by:
$F=q(E+v\times B)=0$
This requires the electric force $F_e=qE$ to be equal in magnitude and opposite in direction to the magnetic force $F_m=q(v\times B)$.

2. Direction of the Electric Field:
Since the magnetic force is perpendicular to both the velocity vector $v$ and the magnetic field vector $B$ (by the definition of the cross product), the balancing electric force must also act along this perpendicular direction. Thus, the electric field $E$ must be perpendicular to the direction of motion $v$ and perpendicular to the magnetic field $B$.

3. Magnitude of the Electric Field:
Equating the magnitudes of the electric and magnetic forces:
$qE=qvB\Rightarrow E=vB$

4. Calculation:
Given parameters:
- Speed of light, $c=3\times {10}^8\text{m s}{-1}^{}$
- Velocity of the electron, $v=\frac{c}{100}=3\times {10}^6\text{m s}{-1}^{}$
- Magnetic field, $B=9\times {10}^{-4}\text{T}$
Substituting these values to calculate the magnitude of $E$:
$E=(3\times {10}^6)\times (9\times {10}^{-4})$
$E=27\times {10}^2\text{V m}{-1}^{}$

Therefore, the electric field $E$ must be perpendicular to $B$ and its magnitude is $27\times {10}^2\text{V m}{-1}^{}$.