In a chamber, a uniform magnetic field of 6.5 G (1 G = 10^{–4} T) is maintained. An electron is shot into the field with a speed of 4.8 ×${10}^{6}$ m s^{–1} normal to the field. the radius of the circular orbit is:

(e = 1.6 × 10^{–19} C, ${m}_{e}$ = 9.1×10^{–31} kg)

$\left(1\right)4.2cm\phantom{\rule{0ex}{0ex}}\left(2\right)5.3cm\phantom{\rule{0ex}{0ex}}\left(3\right)4.7cm\phantom{\rule{0ex}{0ex}}\left(4\right)5.2cm$

73%

Subtopic: Lorentz Force |

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A circular coil of 30 turns and a radius of 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60° with the normal of the coil. What will be the magnitude of the counter-torque that must be applied to prevent the coil from turning?

(1) 7.12 N m

(2) 3.13 N m

(3) 6.50 N m

(4) 4.44 N m

73%

Subtopic: Current Carrying Loop: Force & Torque |

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Two concentric circular coils X and Y of radii 16 cm and 10 cm, respectively, lie in the same vertical plane containing the north to south direction. Coil X has 20 turns and carries a current of 16 A, coil Y has 25 turns and carries a current of 18 A. The sense of the current in X is anticlockwise, and clockwise in Y, for an observer looking at the coils facing west. The magnitude and direction of the net magnetic field due to the coils at their centre is:

$\left(1\right)2\pi \times {10}^{-4}T\left(East\right)\phantom{\rule{0ex}{0ex}}\left(2\right)5\pi \times {10}^{-4}T\left(East\right)\phantom{\rule{0ex}{0ex}}\left(3\right)5\pi \times {10}^{-4}T\left(West\right)\phantom{\rule{0ex}{0ex}}\left(4\right)4\pi \times {10}^{-4}T\left(West\right)$

52%

Subtopic: Magnetic Field due to various cases |

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For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its centre is given by, $B=\frac{{\mu}_{0}I{R}^{2}N}{2{\left({x}^{2}+{R}^{2}\right)}^{{\displaystyle \frac{3}{2}}}}$

The magnetic field at the centre of the coil is:

$\left(1\right)\frac{{\mu}_{0}IN}{R}\phantom{\rule{0ex}{0ex}}\left(2\right)\frac{2{\mu}_{0}IN}{R}\phantom{\rule{0ex}{0ex}}\left(3\right)0\phantom{\rule{0ex}{0ex}}\left(4\right)\frac{{\mu}_{0}IN}{2R}$

73%

Subtopic: Magnetic Field due to various cases |

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A toroid has a core (non-ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which 3500 turns of a wire are wound. If the current in the wire is 11 A, the magnetic field inside the core of the toroid is:

$\left(1\right)3\times {10}^{-2}T\phantom{\rule{0ex}{0ex}}\left(2\right)0\phantom{\rule{0ex}{0ex}}\left(3\right)2\times {10}^{-3}T\phantom{\rule{0ex}{0ex}}\left(4\right)1\times {10}^{-2}T$

Subtopic: Magnetic Field due to various cases |

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An electron emitted by a heated cathode and accelerated through a potential difference of 2.0 kV, enters a region with a uniform magnetic field of 0.15 T. if the field is transverse to its initial velocity, the radius of the circular path is:

(1) 2.10 mm

(2) 0.11 mm

(3) 1.01 mm

(4) 0.12 mm

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Subtopic: Lorentz Force |

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A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?

$\left(1\right)3.14\times {10}^{-4}T\phantom{\rule{0ex}{0ex}}\left(2\right)2.12\times {10}^{-4}T\phantom{\rule{0ex}{0ex}}\left(3\right)1.41\times {10}^{-4}T\phantom{\rule{0ex}{0ex}}\left(4\right)2.01\times {10}^{-4}T$

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Subtopic: Magnetic Field due to various cases |

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A long straight wire carries a current of 35 A. The magnitude of the magnetic field at a point 20 cm from the wire is:

$\left(1\right)3.5\times {10}^{-6}T\phantom{\rule{0ex}{0ex}}\left(2\right)3.5\times {10}^{-5}T\phantom{\rule{0ex}{0ex}}\left(3\right)4.5\times {10}^{-6}T\phantom{\rule{0ex}{0ex}}\left(4\right)4.5\times {10}^{-5}T$

78%

Subtopic: Magnetic Field due to various cases |

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A long straight wire in the horizontal plane carries a current of 50 A in the north to south direction. The magnitude and direction of the magnetic field at a point 2.5 m east of the wire is:

$\left(1\right)4\times {10}^{-6}Tverticallyupward\phantom{\rule{0ex}{0ex}}\left(2\right)4\times {10}^{-6}Tverticallydownward\phantom{\rule{0ex}{0ex}}\left(3\right)3\times {10}^{-6}Tverticallyupward\phantom{\rule{0ex}{0ex}}\left(4\right)3\times {10}^{-6}Tverticallydownward$

64%

Subtopic: Magnetic Field due to various cases |

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A horizontal overhead power line carries a current of 90 A in the east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?

$\left(1\right)1.2\times {10}^{-5}Ttowardsnorth\phantom{\rule{0ex}{0ex}}\left(2\right)2.1\times {10}^{-5}Ttowardssouth\phantom{\rule{0ex}{0ex}}\left(3\right)1.2\times {10}^{-5}Ttowardssouth\phantom{\rule{0ex}{0ex}}\left(4\right)2.1\times {10}^{-5}Ttowardsnorth$

57%

Subtopic: Magnetic Field due to various cases |

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