A conducting rod PQ of length L = 1.0 , m is | vedclass

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Question

(A) The motional electromotive force $(EMF)$ induced in the rod $PQ$ is given by $\varepsilon = Bvl$.
Substituting the given values: $\varepsilon = 4

Vedclass · Accepted Answer

$q_A = + 80 \, \mu C$ and $q_B = - 80 \, \mu C$

Vedclass · Answer

(A) The motional electromotive force $(EMF)$ induced in the rod $PQ$ is given by $\varepsilon = Bvl$.
Substituting the given values: $\varepsilon = 4.0 \, T 	imes 2 \, m/s 	imes 1.0 \, m = 8.0 \, V$.
The charge stored in the capacitor is $Q = C\varepsilon$.
$Q = 10 \, \mu F 	imes 8.0 \, V = 80 \, \mu C$.
According to Fleming's right-hand rule,the induced current flows from $Q$ to $P$ inside the rod. Thus,$P$ acts as the positive terminal and $Q$ acts as the negative terminal.
Following the circuit,the plate $A$ of the capacitor is connected to the higher potential terminal $P$,and plate $B$ is connected to the lower potential terminal $Q$.
Therefore,plate $A$ acquires a positive charge $q_A = + 80 \, \mu C$ and plate $B$ acquires a negative charge $q_B = - 80 \, \mu C$.

$A$ conducting rod $PQ$ of length $L = 1.0 \, m$ is moving with a uniform speed $v = 2 \, m/s$ in a uniform magnetic field $B = 4.0 \, T$ directed into the paper. $A$ capacitor of capacity $C = 10 \, \mu F$ is connected as shown in the figure. Then:

Similar Questions

$A$ wheel with $10$ metallic spokes,each $0.5 \ m$ long,is rotated with a speed of $120 \ rev/min$ in a plane normal to the horizontal component of Earth's magnetic field $H_E$ at a place. If $H_E = 0.4 \ G$ at the place,then the induced emf is: $(1 \ G = 10^{-4} \ T)$

Find the current in the wire for the configuration shown in the figure. Wire $PQ$ has negligible resistance. $\vec{B}$ is the magnetic field coming out of the paper. $\theta$ is a fixed angle made by $PQ$ as it travels smoothly over two conducting parallel wires separated by a distance $d$.

$A$ wheel with $20$ metallic spokes,each $1 \,m$ long,is rotated with a speed of $120 \,rpm$ in a plane perpendicular to a magnetic field of $0.4 \,G$. The induced emf between the axle and the rim of the wheel will be $\left(1 \;G = 10^{-4} \;T \right)$.

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