A circular current carrying coil has a radius R. The distance from the centre of the coil, on the axis, where B will be  18  of its value at the centre of the coil is | Moving Charges And Magnetism

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 Multiple Choice QuestionsMultiple Choice Questions

51.

A galvanometer acting as a voltmeter should have

  • low resistance in series with its coil

  • low resistance in parallel with its coil

  • high resistance in series with its coil

  • high resistance in parallel with its coil


52.

Magnetic field at point O will be

 

  • μo I2R interior

  • μoI2R exterior

  • μo I2R1 - lπ interior

  • μoI2R 1 + Iπ exterior


53.

If the electric flux entering and leaving an enclosed surface respectively are ϕ1 and ϕ2 , the electric charge inside the surface will be

  • ϕ2 - ϕ1εo

  • ϕ2 + ϕ1εo

  • ϕ1 - ϕ2ε0

  • ε0 ϕ1 +  ϕ2 


54.

Two long straight wires, each carrying an electric current of 5 A, are kept parallel to each other at a separation of 2.5 cm. Find the magnitude of the magnetic force experiment by 10 cm of a wire.

  • 4.0 × 10-4 N

  • 3.5 × 10-6 N

  • 2.0 × 10-5 N

  • 2.0 × 10-9 N


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55.

The parts of two concentric circular arcs joined by two radial lines and carries current i. The arcs subtend an angle 0 at the centre of the circle. The magnetic field at the centre O, is

  • μ0 i b - a  θ4π ab

  • μ0i b - aπ - θ 

  • μ0i  b - a θπab

  • μ0i a - b2πab


56.

A square loop is made by a uniform conductor wire as shown in figure

     

The net magnetic field at the centre of the loop if side length of the square is a

  • μ0 i2 a

  • zero

  • μ0 i2a2

  • None of these


57.

A thin bar magnet of length 2 L is bent at the mid-point so that the angle between them is 60°. The new length of the magnet is

  • L/2√3

  • √3 L/2

  • L

  • 2 L/3


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58.

A circular current carrying coil has a radius R. The distance from the centre of the coil, on the axis, where B will be  18  of its value at the centre of the coil is

  • R3

  • 3 R

  • 3 R

  • 2 R3


B.

3 R

        B = μ0 Ni R22 R2 + x2 32

           = μ0N i2 R 11 + x2R232

Let   BCμ0 Ni2R

       B =  BC 11 + x2R232    

⇒      BC8 = BC1 +x2R232                           B = BC8

⇒      1 + x2R2 32 = 8

⇒     1 + x2R2 = 4

⇒       x2R2  = 3

⇒       x = R 3 


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59.

A current carrying loop is placed in a uniform magnetic field in four different orientations I, II, III and IV as shown in figure. Arrange them in decreasing order of potential energy

  • l > lll> ll > lV

  • l > ll > lll > lV

  • l > lV > ll > lll

  • lll > lV > l > ll


60.

A current I is flowing through the loop. The direction of the current and the shape of the loop are as shown in the figure. The magnetic field at the centre of the loop is μ0I/R times ( MA = R, MB = 2R, ∠DMA = 90o )

                     

  • 5/16, but out of the plane of the paper.

  • 5/16, but into the plane of the paper.

  • 7/16, but out of the plane of the paper.

  • 7/16, but into the plane of the paper.


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