The current in an L – R circuit builds up to ${\left(3/4\right)}^{\mathrm{th}}$ of its steady state value in 4 seconds. The time constant of this circuit is?

1. $\frac{1}{\ln <mspace\; width="1em"></mspace>2}\sec$                       

2. $\frac{2}{\ln <mspace\; width="1em"></mspace>2}\sec$

3. $\frac{3}{\ln <mspace\; width="1em"></mspace>2}\sec$                       

4. $\frac{4}{\ln <mspace\; width="1em"></mspace>2}\sec$

The magnetic flux through each turn of a 100 turn coil is $\left({\mathrm{t}}^3<mspace\; width="1em"></mspace>–<mspace\; width="1em"></mspace>2\mathrm{t}\right)<mspace\; width="1em"></mspace>\times <mspace\; width="1em"></mspace>{10}^{-3}<mspace\; width="1em"></mspace>\mathrm{Wb}$, where t is in second. The induced emf at t = 2 s is

1. $-4<mspace\; width="1em"></mspace>\mathrm{V}$                       

2. $-1<mspace\; width="1em"></mspace>\mathrm{V}$

3. $+1<mspace\; width="1em"></mspace>\mathrm{V}$                       

4. $+4<mspace\; width="1em"></mspace>\mathrm{V}$

An emf of 15 volt is applied in a circuit containing 5 henry inductance and 10 ohm resistance. The ratio of the currents at time t = $\infty$ and at t = 1 second is?

1.  $\frac{{\mathrm{e}}^{1/2}}{{\mathrm{e}}^{1/2}-1}$                               

2.  $\frac{{\mathrm{e}}^2}{{\mathrm{e}}^2-1}$

3.  $1-{\mathrm{e}}^{-1}$                           

4.  ${\mathrm{e}}^{-1}$

A circular coil of radius 5 cm has 500 turns of a wire. The approximate value of the coefficient of self induction of the coil will be -

1.  $25<mspace\; width="1em"></mspace>\mathrm{mH}$                           

2.  $25\times {10}^{-3}<mspace\; width="1em"></mspace>\mathrm{mH}$

3.  $50\times {10}^{-3}<mspace\; width="1em"></mspace>\mathrm{mH}$               

4.  $50\times {10}^{-3}<mspace\; width="1em"></mspace>\mathrm{H}$

A coil of radius 1 cm and  turns 100 is placed in the middle of a long solenoid of radius 5 cm and having 8 turns/cm. The mutual induction in millihenry will be-

1.  0.0316                   

2.  0.063 

3.  0.105                     

4.  Zero

A copper rod of length 0.19 m is moving parallel to a long wire with a uniform velocity of 10 m/s. The long wire carries 5 ampere current and is perpendicular to the rod. The ends of the rod are at distances 0.01 m and 0.2 m from the wire. The emf induced in the rod will be-

1.  $10<mspace\; width="1em"></mspace>\mathrm{\mu V}$                           

2.  $20<mspace\; width="1em"></mspace>\mathrm{\mu V}$

3.  $30<mspace\; width="1em"></mspace>\mathrm{\mu V}$                           

4.  $40<mspace\; width="1em"></mspace>\mathrm{\mu V}$

Rate of increment of energy in an inductor with time in series LR circuit getting charge with battery of emf E is best represented by – [inductor has initially zero current]

1.                        

2. 

3.                        

4. 

A long solenoid having 1000 turns per cm is carrying alternating current of one ampere peak value. A search coil of area of cross-section $1\times {10}^{-4}<mspace\; width="1em"></mspace>{\mathrm{m}}^2$ and of 20 turns is placed in the middle of the solenoid so that its plane is perpendicular to the axis of the solenoid. The search coil registers a peak voltage $2.5\times {10}^{-2}<mspace\; width="1em"></mspace>\mathrm{V}$. The frequency of the current in the solenoid is -

1.  1.6 per second                   

2.  0.16 per second

3.  15.9 per second                 

4.  15.85 per second

A coil of the area $7<mspace\; width="1em"></mspace>{\mathrm{cm}}^2$ and of 50 turns is kept with its plane normal to a magnetic field B. A resistance of 30 ohms is connected to the resistance-less coil. B is $75{\mathrm{e}}^{-200\mathrm{t}}$ Gauss. The current passing through the resistance at t = 5 ms will be-

1.  $0.64<mspace\; width="1em"></mspace>\mathrm{mA}$                     

2.  $1.05<mspace\; width="1em"></mspace>\mathrm{mA}$ 

3.  $1.75<mspace\; width="1em"></mspace>\mathrm{mA}$                     

4.  $2.60<mspace\; width="1em"></mspace>\mathrm{mA}$

Assertion: When a coil is joined to a cell, the current through the cell will be 10% less than its steady-state value at time $\mathrm{t}<mspace\; width="1em"></mspace>=\mathrm{\tau }<mspace\; width="1em"></mspace>{\log }_{\mathrm{e}}<mspace\; width="1em"></mspace>10$, where $\mathrm{\tau }$ is the time constant of the circuit.

Reason: When a coil is joined to a cell, the growth of current is given by $\mathrm{i}={\mathrm{i}}_0{\mathrm{e}}^{-\mathrm{t}/\mathrm{\tau }}$