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    Magnetic Field And Earths Magnetism
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    Write the expression for the Lorentz force F in vector form.

    Lorentz force is given by,

    straight F with rightwards harpoon with barb upwards on top space equals space q space open parentheses V with rightwards harpoon with barb upwards on top space x space B with rightwards harpoon with barb upwards on top close parentheses
    where, straight v with rightwards harpoon with barb upwards on top is the velocity of the charged particle, and
    straight B with rightwards harpoon with barb upwards on top is the magnetic field.

    Answer

    (i) Write an expression for Biot-Savart’s law in the vector form. Write an expression for the magnetic field at the centre of a circular coil of radius R, with N turns and carrying a current I. 

    (ii) A helium nucleus completes one round of a circle of radius 0.8 m in 2 seconds. Find the magnetic field at the centre of the circle.

    Expression for Biot-Savart’s Law in the vector form,


    Error converting from MathML to accessible text. 
    So, for N turns, 


    open vertical bar straight B with rightwards harpoon with barb upwards on top close vertical bar space equals space fraction numerator mu subscript o space I space N over denominator 2 r end fraction 
    The direction of B is perpendicular to the plane of the coil directed outward.

    ii) As,

     straight B with rightwards harpoon with barb upwards on top space equals space fraction numerator mu subscript o space I over denominator 2 r end fraction space equals space fraction numerator mu subscript o q over denominator 2 r t end fraction space

    Here,

    q = Charge on 2 electrons

       = 2 x 1.6 x 10-19 C

       = 3.2 x 10-19 C.

    t = 2 seconds.

    r = 0.8 m

    B = fraction numerator 4 straight pi space straight x space 10 to the power of negative 7 end exponent space straight x space 3.2 space straight x space 10 to the power of negative 19 end exponent over denominator 2 space straight x space 0.8 space straight x space 2 end fraction 

       = 4π x 10-26 


       = 4 x 3.14 x 10-26 

       = 1256 x 10-25T 
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    Answer
    A magnet makes 12 oscillations per minute in the earth’s magnetic field alone and 15 oscillations per minute when a short magnet with its axis horizontal and its south pole pointing north is placed with its center 20 cm directly above the oscillating magnet. Find the magnetic moment of the magnet. (Given : BH= 0.36 x 10-4 Tesla).
    open parentheses straight t subscript 1 over straight t subscript 2 close parentheses squared space equals space fraction numerator B subscript H space plus space B over denominator B subscript H end fraction rightwards double arrow space B space equals space B subscript H space open parentheses t subscript 1 over t subscript 2 close parentheses squared space space minus space 1 rightwards double arrow space open parentheses t subscript 1 over t subscript 2 close parentheses squared space minus space 1 space equals space B over B subscript H

    where,

    tl = 15 oscillation/sec and t2 = 12 oscillation/sec and

    B    H = 0.36 x 10-4 T

    rightwards double arrow space open parentheses 15 over 12 close parentheses squared space minus space 1 space equals space B over B subscript H rightwards double arrow space fraction numerator 225 space minus space 144 over denominator 144 end fraction space equals space fraction numerator B over denominator 0.36 space x space 10 to the power of negative 4 end exponent end fraction rightwards double arrow space B space equals space fraction numerator 81 space x space 0.36 space x space 10 to the power of negative 4 end exponent over denominator 144 end fraction rightwards double arrow space B space equals space 2.025 space x space 10 to the power of negative 5 end exponent rightwards double arrow space B space equals space fraction numerator mu subscript o over denominator 4 pi end fraction. space fraction numerator 2 M apostrophe over denominator d cubed end fraction space space space space space space M apostrophe space equals space fraction numerator 2 pi over denominator mu subscript o end fraction. space B d cubed space O n space p u t t i n g space t h e space v a l u e s comma space space space space space space M apostrophe space equals space 20.25 space
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    Answer
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    State and prove Tangent law. 

    Suppose a small bar magnet is suspended which acts right angle to each other to uniform magnetic field B  and BH , such that it comes to rest making an angle θ with the direction of field BH 

    According to tangent law

                               B = B    H tan θ

    Where B and BH are two uniform magnetic fields which act at right angle to each other

    on a magnet and θ is the angle which the magnet makes with the direction of BH.

    tan θ = 
    B = BH tan θ

    Torque due to the magnetic field BH, 

                  τ1 = mBH x MN                                 ...(i)

    Torque due to the magnetic field B,

                   τ2 = mB x OM                                ...(ii)

    In equilibrium position, we have

                         τ1 = τ2

    mB x OM = mBH x MN

    B = BH x 
    In right angled 
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    Explain with necessary formula (no derivation) how you will compare the magnetic moments of two magnets by equal distance method using a deflection magnetometer in tan A position.   
    To compare the magnetic moments, tan A position equal distance method will be used.

    One of the magnets of the magnetic moment M1 is placed on one of its arms.

    The readings of both the ends of aluminium pointer are noted.

    The magnet is reversed end to end in the same position and the readings of both the ends of the pointer are again noted.

    The magnet is now taken to the other arm and the readings are taken by keeping the magnet at the same distance d from the centre of the compass box.

    The average value of these eight readings is forced out as θ1.

    The experiment is repeated with the second magnet of Moment M2, keeping the magnet at the distance d from the centre of the compass box.

    The average of the eight readings is noted as θ2.

    Let 2l1 and 2l2 be the lengths of the two magnets. The magnetic fields produced by the magnets of moments M1 and M2 at the centre of the compass box are B1 and B2respectively.

    If BH is the horizontal component of the magnetic field at the place then,

     

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