Moving Charges And Magnetism

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Question
CBSEENPH12039736

A charged oil drop is suspended in a uniform field of 3 × 104 V/m so that it neither falls nor rises. The charge on the drop will be (take the mass of the charge = 9.9 × 10−15 kg and g = 10 m/s2 )

  • 3.3 × 10−18 C

  • 3.2 × 10−18 C

  • 1.6 × 10−18 C

  • 4.8 × 10−18 C

Solution

A.

3.3 × 10−18 C

In steady state, electric force on drop = weight of drop
∴ qE = mg
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Question
CBSEENPH12039703

A charged particle of mass m and charge q travels on a circular path of radius r that is perpendicular to a magnetic field B. The time taken by the particle to complete one revolution is

  • fraction numerator 2 πmq over denominator straight B end fraction
  • fraction numerator 2 πq squared straight B over denominator straight m end fraction
  • fraction numerator 2 πqB over denominator straight m end fraction
  • fraction numerator 2 πm over denominator qB end fraction

Solution

D.

fraction numerator 2 πm over denominator qB end fraction WiredFaculty

Question
CBSEENPH12039723

A current I ampere flows along an infinitely long straight thin-walled tube, then the magnetic induction at any point inside the tube is

  • infinite

  • zero

  • fraction numerator straight mu subscript 0 over denominator 4 straight pi end fraction fraction numerator 2 straight i over denominator straight r end fraction space tesla
  • fraction numerator 2 space straight i over denominator straight r end fraction space Tesla

Solution

B.

zero

Let R be the radius of a long thin cylindrical shell. To calculate the magnetic induction at a distance r (r < R) from the axis of cylinder, a circular shell of radius r is shown:
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Since no current is enclosed in the circle so, from Ampere's circuital law, magnetic induction is zero at every point of the circle. Hence, the magnetic induction at any point inside the infinitely long straight thin-walled tube (cylindrical) is zero.

Question
CBSEENPH12039591

A current loop ABCD is held fixed on the plane of the paper as shown in the figure. The arcs BC (radius = b) and DA (radius = a) of the loop are joined by two straight wires AB andCD. A steady current I is flowing in the loop. Angle made by AB and CD at the origin O is 30º. Another straight thin wire with steady current I1 flowing out of the plane of the paper is kept at the origin.

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The magnitude of the magnetic field (2) due to the loop ABCD at the origin (O) is

  • zero

  • fraction numerator straight mu space left parenthesis straight b minus straight a right parenthesis over denominator 24 ab end fraction
  • fraction numerator straight mu subscript straight o straight I over denominator 4 space straight pi end fraction space open square brackets fraction numerator straight b minus straight a over denominator ab end fraction close square brackets
  • fraction numerator straight mu subscript 0 straight I over denominator 4 straight pi end fraction space open square brackets 2 left parenthesis straight b minus straight a right parenthesis space plus straight pi over 3 left parenthesis straight a plus straight b right parenthesis close square brackets

Solution

B.

fraction numerator straight mu space left parenthesis straight b minus straight a right parenthesis over denominator 24 ab end fraction WiredFaculty

Question
CBSEENPH12039592

A current loop ABCD is held fixed on the plane of the paper as shown in the figure. The arcs BC (radius = b) and DA (radius = a) of the loop are joined by two straight wires AB andCD. A steady current I is flowing in the loop. Angle made by AB and CD at the origin O is 30º. Another straight thin wire with steady current I1 flowing out of the plane of the paper is kept at the origin.

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Due to the presence of the current I1 at the origin


The magnitude of the magnetic field (2) due to the loop ABCD at the origin (O) is

  • The forces on AB and DC are zero

  • The forces on AD and BC are zero

  • The magnitude of the net force on the loop is given by fraction numerator straight mu subscript straight o II subscript 1 over denominator 4 straight pi end fraction space open square brackets 2 left parenthesis straight b minus straight a right parenthesis space plus space straight pi over 3 left parenthesis straight a plus straight b right parenthesis close square brackets

  • The magnitude of the net force on the loop is given by fraction numerator straight mu subscript 0 space II subscript 1 over denominator 24 space ab end fraction space left parenthesis straight b minus straight a right parenthesis

Solution

B.

The forces on AD and BC are zero

The forces on AD and BC are zero because magnetic field due to a straight wire on AD and BC is parallel to elementary length of the loop.

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