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Byju's Answer

Standard XII

Physics

Volume Stress

A capacitor o...

Question

A capacitor of capacitance $C = .015 F$ is connected to parallel conducting rail and a conducting rod of mass m= $100$ g and length $l = 1 m$ start the fall under gravity in vertical plane. A uniform magnetic field of $2 T$ exist in space direction perpendicular to rod as shown in figure. Find acceleration of rod.

6.25 m / s^{2}

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1.69 m / s^{2}

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1.44 m / s^{2}

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2.25 m / s^{2}

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Solution

The correct option is A $6.25 m / s^{2}$
$\begin{matrix} E = B l V Q = C E = B l V C i = \frac{d Q}{d t} = B l C \frac{d V}{d t} \Rightarrow F_{b} = B i l = B . B l C (\frac{d V}{d t}) l = B^{2} l^{2} C (\frac{d V}{d t}) ∴ m g - B^{2} l^{2} C . a = m a \Rightarrow m g - (m + B^{2} l^{2} C) C \Rightarrow a = \frac{m g}{m + B^{2} l^{2} C} = \frac{1}{\frac{100}{1000} + 4 \times 1^{2} \times \frac{15}{1000}} = \frac{1000}{160} = 6.25 m / s^{2} \end{matrix}$

Hence,

option $A$ is correct answer.

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A capacitor of capacitance C=.015F is connected to parallel conducting rail and a conducting rod of mass m=100g and length l=1m start the fall under gravity in vertical plane. A uniform magnetic field of 2T exist in space direction perpendicular to rod as shown in figure. Find acceleration of rod.

The correct option is A 6.25m/s2 E=BlVQ=CE=BlVCi=dQdt=BlCdVdt⇒Fb=Bil=B.BlC(dVdt)l=B2l2C(dVdt)∴mg−B2l2C.a=ma⇒mg−(m+B2l2C)C⇒a=mgm+B2l2C=11001000+4×12×151000=1000160=6.25m/s2 Hence, option A is correct answer.