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(A) The magnetic field $\vec{B}$ inside a long solenoid is directed along its axis.Since the particle is released from rest and moves along the axis o

Vedclass · Accepted Answer

$a = g$

Vedclass · Answer

(A) The magnetic field $\vec{B}$ inside a long solenoid is directed along its axis.Since the particle is released from rest and moves along the axis of the solenoid,its velocity vector $\vec{v}$ is always parallel or anti-parallel to the magnetic field vector $\vec{B}$.The magnetic force on a moving charge is given by $\vec{F}_{B} = Q(\vec{v} 	imes \vec{B})$.Since $\vec{v}$ is parallel to $\vec{B}$,the cross product $\vec{v} 	imes \vec{B} = 0$,hence $\vec{F}_{B} = 0$.The only force acting on the particle is the gravitational force $\vec{F}_{g} = m\vec{g}$ acting downwards.Therefore,the net force $\vec{F}_{net} = m\vec{g}$.According to Newton's second law,$m\vec{a} = m\vec{g}$,which gives $\vec{a} = \vec{g}$.Thus,the acceleration of the particle is $a = g$.

$A$ current-carrying solenoid is placed vertically and a particle of mass $m$ with charge $Q$ is released from rest. The particle moves along the axis of the solenoid. If $g$ is the acceleration due to gravity,then the acceleration $(a)$ of the charged particle will satisfy:

Similar Questions

In the following diagram,there is a straight wire carrying a current $I$. Consider a circular path with radius $R$ near it. If $\vec{B}_T$ is the tangential component of the magnetic field along the circular path,then find the value of the integral $\oint \vec{B}_T \cdot d\vec{l}$.

$A$ long solenoid of $50\, cm$ length having $100$ turns carries a current of $2.5\, A$. The magnetic field at the centre of the solenoid is $...... \times 10^{-5}\, T$. $(\mu_{0} = 4\pi \times 10^{-7}\, T\, m\, A^{-1})$

If $B$ is the magnetic field and $\mu_0$ is the permeability of free space,then the dimensions of $(B / \mu_0)$ are:

The figure shows a long straight wire of circular cross-section (radius $a$) carrying a steady current $I$. The current $I$ is uniformly distributed across this cross-section. Calculate the magnetic field in the region $r < a$ and $r > a$.

The number of turns per unit length of a long solenoid is $10 \, \text{turns/cm}$. If its average radius is $5 \, \text{cm}$ and it carries a current of $10 \, \text{A}$, then the ratio of flux densities obtained at the centre and at the end on the axis will be:

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