Magnetic force on a moving charge

The magnetic force on a charged particle is given by $\vec{F} = q\,\vec{v} \times \vec{B}$ and its magnitude is $F = qvB\sin\theta$ where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$.

Condition for straight line motion with constant speed

For the electron to move in a straight line with constant velocity, the net magnetic force must be zero: $\vec{F}=0$

This happens when $\vec{v} \times \vec{B}=0$ which is possible if $\vec{v}$ is parallel or antiparallel to $\vec{B}$, i.e. $\theta=0^\circ$ or $180^\circ$.

Check Assertion (A)

Assertion says: An electron in a certain region of uniform magnetic field is moving with constant velocity in a straight line path.

This is possible if the electron moves along the magnetic field direction. Then magnetic force is zero, so no deflection occurs.

Hence, Assertion (A) is true.

Check Reason (R)

Reason says: The magnetic field in that region is along the direction of velocity of the electron.

If $\vec{B}$ is along $\vec{v}$, then $\vec{v} \times \vec{B}=0$ so magnetic force is zero and the electron continues in a straight line with constant velocity.

Hence, Reason (R) is true.

Relation between A and R

The reason correctly explains why the electron can move in a straight line with constant velocity in a magnetic field.

Correct option

Therefore, the correct answer is: $\boxed{\text{B}}$

Both (A) and (R) are true and (R) is the correct explanation of (A).

Comparison with stored answer

Stored correct answer: B

My derived answer: B

So, they agree.