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Electromagnetic Induction Test - 76

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Electromagnetic Induction Test - 76
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  • Question 1
    1 / -0
    A 1.2m1.2m wide track is parallel to the magnetic meridian. The verticle component of the earth's magnetic field is 0.50.5 gauss. When a train runs on the rails at a speed of 60Km/hr60Km/hr, then the induced potential difference between the ends of its axle will be __
    Solution

  • Question 2
    1 / -0
    Two cells of emf 2V2V and 1.5V1.5 V with internal resistance 1Ω1 {\Omega} each are connected in parallel similar poles joined together. The combination is connected to an external resistance of 10Ω10{\Omega}. Find the current through the external resistance. 
  • Question 3
    1 / -0
    Two coils A and B have mutual inductance 2×1022\times { 10 }^{ -2 } henry. If the current in the primary is i=5sin(10πt) i=5\sin { \left( 10\pi t \right)  } then the maximum value of e.m.f.induced in coil B is 
    Solution

  • Question 4
    1 / -0
    An electron moves on a staight line path XYXY as shown. The abcd is a coil adjacent to the path of electron. What will be the direction of current, if any,induced in the coil?

    Solution

  • Question 5
    1 / -0
    The magnetic induction at the centre OO in the figure shown is

    Solution
    Correct Answer: Option A  

     
    Hint: Resultant magnetic field is equal to sum of the magnetic field due to the currents in the two semicircles. This can be found using the equation of  magnetic field at the centre of a loop.


    Explanation of Correct Option:


    Step 1: Finding magnetic field due to the two loops.

    Let the magnetic field due to the semicircular loop of radius R1R_{1} be B1B_{1} and magnetic field due to the semicircular loop of radius R2R_{2} be B2B_{2}

    Resultant magnetic field at O=O= Magnetic field due to the semicircular loop of radius R1R_{1}Magnetic field due to the semicircular loop of radius R2R_{2}
                                                      =B1+B2=B_{1}+B_{2}

    Magnetic field at the centre of a current carrying loop =μ0i2 r= \dfrac{\mu_{0}i}{2\ r}

    Magnetic field at the centre of a semicircular current carrying loop =12×μ0i2 r=μ0i4 r=\dfrac{1}{2}\times \dfrac{\mu_{0}i}{2\ r} = \dfrac{\mu_{0}i}{4\ r}

    B1=μ0i4 R1B_{1}=\dfrac{\mu_{0}i}{4\ R_{1}}

    B2=μ0i4 R2B_{2}=-\dfrac{\mu_{0}i}{4\ R_{2}}           (Negative sign since the direction is opposite)


    Step 2: FInding the magnetic field due to horizontal parts
    We know, 
    dB=μi4π(dlr^sinθr2)dB=\dfrac{\mu i}{4\pi}(\dfrac{\vec{dl}\hat{r}sin\theta} {r^{2}})
    Here since the angle is 00^{\circ} and 180180^{\circ}, magnetic field due to the horizontal parts is zero.


    Step 3: Finding magnetic induction at the centre.

    Magnetic induction at the centre O,B=B1+B2O, B= B_{1}+B_{2}
                                                                
                                                                B=μ0i4 R1+μ0i4 R2B=\dfrac{\mu_{0}i}{4\ R_{1}}+-\dfrac{\mu_{0}i}{4\ R_{2}}

                                                                B=μ0i4(1R11R2)B=\dfrac{\mu_{0}i}{4}(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}})


    The magnetic induction at centre in the shown figure is B=μ0i4(1R11R2).B=\dfrac{\mu_{0}i}{4}(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}}).

  • Question 6
    1 / -0
    A wire in the form of a square of side a caries a current i. Then the magnetic induction at the centre of the square wire is( Magnetic permeability of free space=μ  { \mu  }_{ \circ  })

    Solution

  • Question 7
    1 / -0
    A closely wound coil of 100 turns and area of cross section 1cm21{cm}^{2} has a self-inductance 1 mH. The magnetic induction at the centre of the coil, when a current 2A flows in it, will be :

    Solution

  • Question 8
    1 / -0
    A wire carrying current II is tied between points PP and QQ and is in the shape of a circular arch of radius RR due to a uniform magnetic field BB (perpendicular to the plane of the paper, shown by xxx) in the vicinity of the wire. If the wire subtends an angle 2θo2 \theta _o at the centre of the circle (of which it form an arch) then the tension in the wire is :

    Solution

  • Question 9
    1 / -0
    A copper wire is wound on a wooden frame, whose shape is that of an equilateral triangle. If the linear dimension of each side of the frame is increased by a factor of 3, keeping the number of turns of the coil per unit length of the frame the same, then the self inductance of the coil:
    Solution

  • Question 10
    1 / -0
    In the circuit shown, the power factor of the circuit is 35\dfrac { 3 }{ 5 } , Power factor of only RC circuit is 45\dfrac { 4 }{ 5 } , source voltage is 100100 volt and its angular frequency  ω=100rad/sec\omega =100 rad/ sec. RMS current in circuit is A> If inductive reactance is greater then capacitive reactance then the value of self inductance L is 

    Solution

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