A conducting rod PQ of length $1$ m is moving with uniform velocity of $2 m / s$ is a uniform magnetic field of $2 T$ directed into the plane of paper. A capacitor of capacity $c = 10 μ F$ is connected as shown. Then :

q A = + 40 μ C, q B = + 40 μ C

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

q A = + 40 μ C, q B = - 40 μ C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

q A = - 40 μ C, q B = + 40 μ C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

q A = q B = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $q A = + 40 μ C, q B = + 40 μ C$
Motional EMF, $v = B l V$
$= 2 \times 1 \times 2$
$= 4 V$
For a Capacitor,
$Q = C V$
$= ∠ O P F \times 4 V$
$Q = 40 μ C$
The induced any opposes the change that taking place.

Suggest Corrections

Similar questions

Q. 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 μ F

is connected as shown in figure. Then

A conducting rod PQ of length L = 1.0 m is moving with uniform speed v = 2.0 $m s^{- 1}$ in a uniform magnetic field B = 4.0 T directed into the paper. A capacitor of capacity C = 10 $μ$ F is connected as shown in the figure. Then after a long time charge on plates of capacitor is

A conductor A of capacity $4 μ F$ has a charge $20 μ C$ and another condenser B of capacity $10 μ F$ has a charge $40 μ C$ . If they are connected parallel, then:

Q. A neutral conducting sphere of radius

R

contains two spherical cavities of radii

a

and

b

that contains point charges

q_{a}

and

q_{b}

placed at the center of each cavity as shown in the figure below . Match the following table:

$(i) σ_{a}$	$(a) \frac{q_{a} + q_{b}}{4 π R^{2}}$
$(ii) σ_{b}$	$(b) \frac{- q_{a}}{4 π a^{2}}$
$(iii) σ_{R}$	$(c) \frac{- q_{b}}{4 π b^{2}}$

Charges of $40 μ C, - 40 μ C$ and $10 μ C$ are placed in air at the corners A, B and C respectively of an equilateral triangle of side $2 c m$ . The resultant force on the charge at C is :

Join BYJU'S Learning Program

A conducting rod PQ of length 1m is moving with uniform velocity of 2m/s is a uniform magnetic field of 2T directed into the plane of paper. A capacitor of capacity c=10μF is connected as shown. Then :

The correct option is A qA=+40μC,qB=+40μCMotional EMF, v=BlV =2×1×2 =4VFor a Capacitor,Q=CV=∠OPF×4VQ=40μCThe induced any opposes the change that taking place.

A conducting rod PQ of length $1$ m is moving with uniform velocity of $2 m / s$ is a uniform magnetic field of $2 T$ directed into the plane of paper. A capacitor of capacity $c = 10 μ F$ is connected as shown. Then :

The correct option is A $q A = + 40 μ C, q B = + 40 μ C$
Motional EMF, $v = B l V$
$= 2 \times 1 \times 2$
$= 4 V$
For a Capacitor,
$Q = C V$
$= ∠ O P F \times 4 V$
$Q = 40 μ C$
The induced any opposes the change that taking place.