A galvanometer having internal resistance 10 Ω require

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 Multiple Choice QuestionsMultiple Choice Questions

341.

A current loop in a magnetic field

  • experiences a torque whether the field is uniform or non- uniform in all orientations

  • can be in equilibrium in one orientations

  • can be equilibrium in two orientations, both the equilibrium states are unstable

  • can be in equilibrium in two orientations, one stable while the other is unstable


342.

A bar magnet of length l and magnetic dipole moment M is bent in the form of an arc as shown in figure. The new magnetic dipole moment will be

            

  • M

  • 3π M

  • 2π M

  • M2


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343.

A galvanometer having internal resistance 10 Ω requires 0.01 A for a full scale deflection. To convert this galvanometer to a voltmeter of full-scale deflection at 120 V, we need to connect a resistance of

  • 11990 Ω in series

  • 11990 Ω in parallel

  • 12010 Ω in series

  • 12010 Ω in parallel


A.

11990 Ω in series

Resistance R= Vig - G = 1200.01 - 10                     = 12000 - 10                  R = 11990 Ω

To convert a galvanometer into a voltmeter, high resistance should connect in series.


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344.

An electron in a circular orbit ofradius 0.05 nm performs 1016 revolutions per second. The magnetic moment due to this rotation of electron is (in Am2)

  • 2.16 × 10-23

  • 3.21 × 10-22

  • 3.21 × 10-24

  • 1.26 × 10-23


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345.

A very small circular loop of radius a is initially (at t = 0) coplanar and concentric with a much larger fixed circular loop of radius b. A constant current I flows in the larger loop. The smaller loop is rotated with a constant angular speed ω about the common diameter. The emf induced in the smaller loop as a function of time t is

  • πa2μ0I2b ω cos (ωt)

  • πa2μ0I2b ω sin ω2t2

  • πa2μ0I2b ω sin ωt

  • πa2μ0I2b ω sin2 ωt


346.

A proton of mass m and charge q is moving in a plane with kinetic energy E. If there exists a uniform magnetic field B, perpendicular to the plane of the motion, the proton will move in a circular path of radius

  • 2EmqB

  • 2EmqB

  • Em2qB

  • 2EqmB


347.

A long conducting wire carrying a current I is bent at 120° (see figure). The magnetic field B at a point P on the right bisector of bending angle at a distance d from the bend is (µ0 is the permeability of free space)

           

  • 3μ0I2πd

  • μ0I2πd

  • μ0I3 πd

  • 3 μ0I2 πd


348.

A stream of electrons and protons are directed towards a narrow slit in a screen (see figure). The intervening region has a uniform electric field E (vertically downwards) and a uniform magnetic field B (out of the plane of the figure) as shown. Then

            

  • electrons and protons with speed EB will pass through the slit

  • protons with speed EB will pass through the slit, electron of the same speed will not

  • neither electrons nor protons will go through the slit irrespective of their speed

  • electrons will always be deflected upwards irrespective of their speed


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349.

Two particles, A and B, having equal charges, after being accelerated through the same potential difference enter into a region of uniform magnetic field and the particles describe circular paths of radii R1 and R2, respectively. The ratio of the masses of A and B is

  • R1/R2

  • R1 / R2

  • (R1 / R2)2

  • (R2 / R1)2


350.

A straight conductor 0.1 m long moves in a uniform magnetic field 0.1 T. The velocity of the conductor is 15 m/s and is directed perpendicular to the field. The emf induced between the two ends of the conductor is

  • 0.10 V

  • 0.15 V

  • 1.50 V

  • 15.00 V


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