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This is the second of our many quizzes on Solenoid. Click on this to access more such competitive quizzes on Electrostatics and Magnetism.

Solenoid Quiz 2

physics

A huge solenoid is fitted inside a long tunnel of number of turns per unit length $2200$ and current flowing through it is $20\,A$. If the amount of magnetic induction produced by solenoid is given to a loop of area $5{{m}^{2}}$. The magnetic flux in the loop will be 

 The number of terms per unit length and the current flowing through a long solenoid decide the magnetic field at that point. The angle between the point and the ends of a short-length solenoid must be considered.

  $\text{Magnetic induction due to a long solenoid is given by,}\\\overrightarrow{B}={{\mu }_{\circ }}nI\\\Rightarrow \overrightarrow{B}=4\pi \times {{10}^{-7}}\times 2200\times 20=5.52\times {{10}^{-2}}T\\\text{Now magnetic flux through a given area is given by,}\\\phi =\overrightarrow{B}A\\\Rightarrow \phi =5.52\times {{10}^{-2}}\times 5\\\therefore \phi =0.276\,Wb$

One day a high school student went to buy some solenoid for his assignment and he bought multiple solenoids of different numbers of turns and total length. Now he is measuring the relation between magnetic induction produced in solenoid with respect to length of solenoid. The correct graph will be- 

Due to inverse proportionality magnetic induction will decrease with increase in length and graph will not be a straight line. 

  $\text{Magnetic induction produced in a solenoid is given by,}\\B={{\mu }_{\circ }}nI={{\mu }_{\circ }}\dfrac{N}{L}I\\\Rightarrow B\propto \dfrac{1}{L} \\\text{So the magnetic induction is inversely proportional to the length of solenoid.} \\ \text{Hence magnetic induction will decrease with increase in length.}$

A kid is playing with two same solenoids $A$ and $B$, both having the same amount of magnetic induction in it equal to $B$. Now he can add some material in solenoid to increase magnetic induction, so he added medium 1 in solenoid $A$ and medium 2 in solenoid $B$. The ratio of new magnetic induction in solenoid $A$ to $B$ will be- $[\text{Relative permeability of medium A is 2000 and that of 2 is 2250}]$ 

Relative permeability can be termed as the ratio of actual permeability and that of permeability of free medium which  causes change in magnetic field.	

 $\text{Magnetic induction in a solenoid will be given by,}\\B={{\mu }_{\circ }}nI\\\text{If we add some material in it so magnetic fields in both solenoid will depend on relative permeability of materials so,}\\B={{\mu }_{r}}nI\\\text{The magnetic induction in A and B will be,}\\{{B}_{A}}=2000\times {{\mu }_{\circ }}nI\\\Rightarrow {{B}_{B}}=2250\times {{\mu }_{\circ }}nI\\\text{Now the ratio of magnetic In A to B,}\\\therefore \dfrac{{{B}_{A}}}{{{B}_{B}}}=\dfrac{2000}{2250}=0.88$

 If a solenoid having finite number of turns per unit length $20$ and current induced is $10\,A$. the magnetic field at a point inside the solenoid having an angle of ${{30}^{\circ }}$ and ${{60}^{\circ }}$ with the end points is given by-<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/d3dd2c6c-d81b-4f9d-8346-f6643b7e4c981644514899036243201.png"/>

 Magnetic field inside a solenoid can be termed as a sum of magnetic field due to individual loops in a solenoid. 

 $\text{Magnetic induction in a finite solenoid will be given by,}\\B=\dfrac{{{\mu }_{\circ }}nI}{2}\left[ \sin ({{\theta }_{1}})+\sin ({{\theta }_{2}}) \right]\\\Rightarrow B=\dfrac{4\pi \times 20\times 10\times {{10}^{-7}}}{2}\left[ \sin ({{60}^{\circ }})+\sin ({{30}^{\circ }}) \right]\\\Rightarrow B=400\pi \times {{10}^{-7}}\left[ \dfrac{\sqrt{3}}{2}+\dfrac{1}{2} \right]\\\Rightarrow B =200\pi \times {{10}^{-7}}\left[ \sqrt{3}+1 \right]\\\therefore B =1.715\times {{10}^{-4}}\,T$

The total power in a current carrying wire of resistance $10\Omega $ is $40$ Watt and the current produced in the wire is due to the long solenoid of number of turns per unit length equal to $250$.The magnetic induction produced by solenoid will be- 

 The length of solenoid is much larger than its diameter so we can neglect distorted fields at ends.

 $\text{Power is given by,}\\P={{I}^{2}}R\\\Rightarrow 40={{I}^{2}}\times 10\\\Rightarrow I=2\\\text{Now magnetic field inside a solenoid is given by,}\\B={{\mu }_{\circ }}nI\\\Rightarrow B =4\pi \times {{10}^{-7}}\times 250\times 2\\\therefore B =6.28\times {{10}^{-4}}\,T$