Physics M.C.Q (1 Marks)

2. The end A becomes positively charged.

Explanation:

The end 'A' becomes, positively charged. Because magnetic field exerts an average Force $\overrightarrow{\text{F}}_0=\text{q}\vec{\text{v}}\times\vec{\text{B} }$ n each free electron where q = 1. 6 × 10¹⁹C is the charge on the electron. This Force is towards AB and hence the free electrons will move towareds B. Negative charge is accumulated at 'B' and positive charge appears at A.

1. Both A and B are true.

Explanation:

Þ An emf con be induced by moving a condcutor in a magnetic field.

$\hat{\text{I}}=\text{Bvl}$

Þ An emf can be induced by charging the magnetic field.

$\in=\frac{-\text{d}\phi}{\text{dt}} \phi\rightarrow\text{flux}$

1. The charge on C long after t = 0 is $\in\text{C}.$

2. The current in L just before t = t₀ is $\frac{\in}{\text{R}}.$

Explanation:

þ A long time after capacitor is fully charged is equal to

$\text{Q}=\text{CV}=\text{C}\hat{\text{I}}$

$\text{Q}=\text{C}\hat{\text{I}}\Big(1-\text{e}^{-\frac{\text{t}}{\text{t}}}\Big)$

þ The current in 'L' just before t = t₀ is

$\text{i}=\frac{\hat{\text{I}}}{\text{R}}\Big(1-\text{e}^{-\frac{\text{t}}{\text{t}}}=\frac{\hat{\text{I}}}{\text{R}}\Big)$

2. Rate of Joule heating if the same current goes through them.

4. Time constant if one solenoid is connected to one battery and the other is connected to another battery.

Explanation:

$\text{R}=\frac{\rho\text{l}}{\text{A}}$

A - Crossectional Area

Thick wire "A" is large than thin wire.

R_thick wire < R_thin wire

þ time constant

þ $\tau=\frac{\text{L}}{\text{R}}$

þ Power dissipatedin Heating $=\text{I}^2\text{R}$

4. None of these.

​​​​​​​Explanation:

Potential difference appears across the two ends $=\text{Bvl}$

$\text{v}^\hat{}\text{B}, \ \text{v}^\hat{}\text{I}, \ \text{I}^\hat{}\text{B}$

3. Move away from the solenoid.

Explanation:

$\text{e}=-\text{L}\frac{\text{di}}{\text{dt}}$

Current flow in the CKt is clock wise direction, due to Mutual Induction current flow in the loop anti clockwise direction. The net force applied on the loop in east direction. So we can say that the ring will move away from the solenoid.

1. $\text{zero}$

Explanation:

Emf at both end is same $=\frac{1}{8}\text{Bwl}^2$

So the potential difference between the two ends of therod is zero.

4. $\text{Zero.}$

Explanation:

Induced emf is AB is Bvl and Induced emf is DC is also Bvl.

Net emf in the closed circuit (loop) is zero.

So induced current in the loop is zero.

1. $\frac{1}{\text{RC}}$

2. $\frac{\text{R}}{\text{L}}$

3. $\frac{1}{\sqrt{\text{LC}}}$

Explanation:

þ Time constant t = RC in RC circuit

frequency $=\frac{1}{\tau}=\frac{1}{\text{RC}} \ ...(\text{i})$

þ Time constant in LR circuit is $\tau=\frac{\text{L}}{\text{RC}}$

frequency $\frac{1}{\tau}=\frac{\text{R}}{\text{L}} \ ...(\text{ii})$

þ eq. (i) & (ii) multiply

frequency $=\frac{1}{\text{LC}}$

frequency $=\frac{1}{\sqrt{2\text{C}}}$

Explanation:

$\in=\text{Bvl}$

Power $=\frac{\text{v}^2}{\text{R}}=\frac{\in^2}{\text{R}}=\frac{\text{B}^2\text{v}^2\text{l}^2}{\text{R}}$

P → Power

v → velocity

B → Magnetic field

Here, $\text{P}\times\text{v}^2$

2. $\frac{1}{8}\omega\text{Bl}^2$

Explanation:

Take a small element dx at a distance of 'x' centre

 $\text{Þ}\text{d}\in\int_{0}^{\frac{1}{2}}\text{B}\omega\text{x}\text{dx}=\frac{\text{B}\omega\text{x}^2}{2}\Big|_{0}^{\frac{1}{2}}$

$\in=\frac{1}{8}\omega\text{Bl}^2$