1. ⇒ (MHT CET 2023 11th May Morning Shift )

An electron is projected along the axis of circular conductor carrying current I. Electron will experience

A. no force.

B. a force along the axis.

C. a force at angle $30^{\circ}$ with the axis.

D. a force perpendicular to axis.

Correct Option is (A)

The magnetic force experienced by the electron, $F = qvB \sin θ$

As electron is projected along the axis, $θ = 0$

$∴ F = 0$

2. ⇒ (MHT CET 2023 10th May Morning Shift )

Electron of mass ' $m$ ' and charge ' $q$ ' is travelling with speed ' $v$ ' along a circular path of radius ' $R$ ', at right angles to a uniform magnetic field of intensity ' $B$ '. If the speed of the electron is halved and the magnetic field is doubled, the resulting path would have radius

A. $4 R$

B. $2 R$

C. $\frac{R}{2}$

D. $\frac{R}{4}$

Correct Option is (D)

From cyclotron motion and uniform circular motion,

$i.e., q v B \sin 90^{\circ} = \frac{{mv}^{2}}{R}$

$∴ R = \frac{mv}{qB}$

Given: $v^{'} = \frac{v}{2}$ and $B^{'} = 2 B$

$\begin{aligned} ∴ R^{'} & = \frac{{mv}^{'}}{{qB}^{'}} \\ = \frac{m \frac{v}{2}}{q 2 B} = \frac{1}{4} \frac{mv}{qB} \\ \Rightarrow R^{'} & = \frac{1}{4} R \end{aligned}$

3. ⇒ (MHT CET 2021 21th September Morning Shift )

A charge moves with velocity ' $V$ ' through electric field $(E)$ as well as magnetic field (B). then the force acting on it is

A. $q (\vec{B} \times \vec{V})$

B. $q (\vec{V} \times \vec{B})$

C. $q \vec{E} + q (\vec{V} \times \vec{B})$

D. $q (\vec{E} \times \vec{V})$

Correct Option is (C)

Force due to electric field ${\vec{F}}_{e} = q \vec{E}$

Force due to magnetic field ${\vec{F}}_{m} = q (\vec{v} \times \vec{B})$

$∴$ Total force $\vec{F} = q \vec{E} + q (\vec{v} \times \vec{B})$

⇒Magnetic Field due to Electric Current