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 A conductor lies along the z- axis  at   $-1.5\leq z<1.5m$   and carries a fixed current of 10.0A in -a_z direction (see figure) . For a field   $ B=3.0\times 10^{-4}e^{-0.2x}a_{y}T,$ find the power required to move the conductor  at constant  speed to x=2.0m, y=0 in   $5\times 10^{-3}s$ . Assume parallel motion along the x-axis

Answer:

Explanation:

 When force exerted on a current carrying conductor

 F_ext=BIL

average power= work done /time taken

 $P=\frac{1}{t}\int_{0}^{2} F_{ext}.dx$

    $=\frac{1}{t}\int_{0}^{2} B(x)IL.dx$

=$\frac{1}{5\times 10^{-3}}\int_{0}^{2} 3\times 10^{-4}e^{-0.2x}\times10\times3 dx$

= $9[1-e^{-0.4}]$

=   $9\left[1-\frac{1}{e^{0.4}}\right]=2.967=2.97W$

 In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100W, 5 fans of 80 W and 1 heater of 1 kW. The voltage of the electric mains is 220V. The minimum  capacity of the main fuse of the building will be

Answer:

Explanation:

 Total power (P) consumed  

                            $=(15\times40)+(5\times100)+(5\times80)+(1\times1000)$

                          =2500W

 As we know,

 power i.e, P=VI

$\Rightarrow $              $ I= \frac{2500}{220}A$

  $= \frac{125}{11}=11.3 A$

                 Minimum capacity should be 12 A

A parallel plate capacitor is made of two circular plates separated by a distance of 5 mm and with a dielectric of dielectric constant  2.2 between them. When the electric field in the dielectric is   $3\times 10^{4}V/m$ , the charge density of the positive plate will be close to

Answer:

Explanation:

 When free space  between parallel plates of capacitor

      $E=\frac{\sigma}{\epsilon_{0}}$

 when dielectric is introduced between parallel plates of capacitor ,    $E'=\frac{\sigma}{k\epsilon_{0}}$

 Electric field inside dielectric

    $\frac{\sigma}{k\epsilon_{0}}=3\times 10^{4}$

 where, k= dielectric   constant  of medium=2.2

  $\epsilon_{0}$= permitivity of free space

   $=8.85\times 10^{-12}$

 $\Rightarrow$       $\sigma=2.2\times8.85\times 10^{-12}\times3 \times 10^{4}$

  $=6.6\times8.85\times 10^{-8}$

  $=5.841\times 10^{-7}$

 =  $6\times 10^{-7}C/m^{2}$

 Assume that  an electric field   $E=30x^{2}\hat{i}$  exists in space . Then, the potential difference V_A-V_O, where V₀ is the potential at the origin  and V_A the potential at x=2 m is

Answer:

Explanation:

 As we know, potential  difference  V_A-V₀ is

   dV=-Edx

$\int_{V_{0}}^{V_{A}} dV=-\int_{0}^{2} 30x^{2}dx$

$V_{A}-V_{0}=-30\times\left[\frac{x^{3}}{3}\right]_0^2$

   = $-10\times[2^{3}-(0)^{3}]$

 = $-10\times 8=-80J$

A pipe of length 85 cm is closed from one end. Find the number of possible  natural oscillations  of air column in the pipe whose frequencies lie below 1250 Hz . The velocity of sound in air is 340 m/s.

Answer:

Explanation:

For closed organ pipe 

     = $\frac{(2n+1)v}{4l}$         [n=0,1,2,...]

$\frac{(2n+1)v}{4l}<1250$

  $(2n+1)<1250\times\frac{4\times0.85}{340}$

  $(2n+1)<1.25\Rightarrow 2n<11.50$

  n<5.25

 n=0,1,2,3,...5

 we have 6 possibilities

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