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 Multiple Choice QuestionsMultiple Choice Questions

331.

A straight conductor of length I carrying a current I, is bent in the form of a semicircle. The magnetic field (in tesla) at the centre of the semicircle is

  • π2Il × 10-7

  • πIl × 10-7

  • πIl2 × 10-7

  • πI2l × 10-7


332.

Which of the following relation represents Biot-Savart's law ?

  • dB  = μ04π dl × rr

  • dB  = μ04π dl × rr3

  • dB  = μ04π dl × rr3

  • dB  = μ04π dl × rr4


 Multiple Choice QuestionsShort Answer Type

333.

A 50 cm long conductor AB moves with a speed 4 m/s in a magnetic field B = 0.01 Wb/m2 as shown. Find the emf generated and power delivered if resistance of the circuit is 0.1 Ω.


334.

An electron is moving with velocity 2 i^ + 2 j^ m/s in an electric field of intensity E = i^ + 2 j^ - 8 k^ volt/m and a magnetic field of B = 2 j^ + 3 k^ tesla. Find the magnitude of force on the electron.


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 Multiple Choice QuestionsMultiple Choice Questions

335.

The ratio of magnetic field and magnetic moment at the centre of a current carrying circular loop is x. When both the current and radius is doubled the ratio will be

  • x/8

  • x/4

  • x/2

  • 2x


336.

Current through ABC and A' B'C' is I. What is the magnetic field at P ? BP = PB' = r (Here C' B' PBC are collinear)

  • B = 14π 2Ir

  • B = μ04π 2Ir

  • B = μ04π Ir

  • Zero


337.

The magnetic field at the point of intersection of diagonals of a square wire loop of side L carrying a current I is

  • μ0IπL

  • 2μ0IπL

  • 2 μ0IπL

  • 22 μ0IπL


338.

A straight wire of length 2 m carries a current of 10 A. If this wire is placed in a uniform magnetic field of 0.15 T making an angle of 45° with the magnetic field, the applied force on the wire will be

  • 1.5 N

  • 3 N

  • 32 N

  • 32 N


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339.

A magnetic needle is placed in a uniform magnetic field and is aligned with the field. The needle is now rotated by an angle of 60° and the work done is W. The torque on the magnetic needle at this position is

  • 23 W

  • 3 W

  • 32 W

  • 34 W


B.

3 W

Given, work done = W and θ = 60°

We know that  W = MB (1 − cos θ)

                      W = MB (1 − cos 60°)

                      W = MB2

Hence,           ζ = MBsin 60° = 3 W


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340.

When a proton is released from rest in a room, it starts with an initial acceleration a0 towards west. When it is projected towards north with a speed v0 it moves with an initial acceleration 3a0 towards west. The electric and magnetic fields in the room are

  • ma0e west, 2ma0ev0 up

  • ma0e west, 2ma0ev0 down

  • ma0e east, 3ma0ev0 up

  • ma0e east, 3ma0ev0 down


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