Question

A Charged sphere of mass $m$ and charge $q$ starts sliding from rest on a vertical fixed circular track of radius $R$ from the position as shown in figure. There exists a constant magnetic field in horizontal direction as shown in figure. The maximum force exerted by track on the sphere is:

• A. $3mg+qB\sqrt{2gR}$
• B. $mg+3qB\sqrt{gR}$
• C. $2mg+qB\sqrt{gR}$
• D. $2mg+3qB\sqrt{2gR}$

Answer 1

The correct option is A $3mg+qB\sqrt{2gR}$

Using Relation for magnetic force, ${\overrightarrow{F}}_m=q(\overrightarrow{v}\times \overrightarrow{B})$

The direction of $\overrightarrow{v}$ is always tangential to the track,

Thus , $(\overrightarrow{v}\times \overrightarrow{B})$ gives the direction of ${\overrightarrow{F}}_m$ along the normal direction shown below in FBD:




Using eq. for circular dynamics:

$F_c=\frac{m{v}^2}{R}$

$N-mg\sin \theta -F_m=\frac{m{v}^2}{R}$ ....... (i)

Here, $F_m=qvB\sin {90}^{\circ }=qvB$

$\Rightarrow N-mg\sin \theta -qvB=\frac{m{v}^2}{R}$

$N=\frac{m{v}^2}{R}+qvB+mg\sin \theta$ ...... (ii)

The normal reaction will be maximum when $mg\sin \theta$ is maximum or $\theta ={90}^{\circ }$

Hence at $\theta ={90}^{\circ }$ the vertical displacement of body is ${}^{\prime }{R}^{\prime }$ downwards.

From conservation of mechanical energy,

loss in P.E = Gain in K.E

$mgR=\frac{1}{2}m{v}^2$

$v=\sqrt{2gR}$....(iii)

From Eq.(ii) & (iii) we get,

$N_{max}=\frac{m(\sqrt{2gR}{)}^2}{R}+q\left(\sqrt{2gR}\right)B+mg$

$N_{max}=\frac{2mgR}{R}+mg+qB\sqrt{2gR}$

$\therefore N_{max}=3mg+qB\sqrt{2gR}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option (a) is the correct answer.

Why this question ? Tip:In circular motion, $\overrightarrow{v}\text{is always tangential along circular path and}\overrightarrow{v}\times \overrightarrow{B}$ gives that ${\overrightarrow{F}}_m$. Further, apply the equation of dynamics at any angle $\theta$ w.r.t horizontal (ralease line) and $v$ can be obtained using mechanical energy conservation.