An electron in the beam of a TV picture tube is accelerated by a PD of 2.00kv. Then it passes through a region of transverse magnetic field where it moves in a circular Arc with radius 0.180m. the magnitude of the field is

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Arun · Answer

Dear student I am getting my answer as 8.38 * 10^-4 T Hope it helps Regards Arun (askIITians forum expert)

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An electron in the beam of a TV picture tube is accelerated by a PD of 2.00kv. Then it passes through a region of transverse magnetic field where it moves in a circular Arc with radius 0.180m. the magnitude of the field is

An electron in the beam of a TV picture tube is accelerated by a PD of 2.00kv. Then it passes through a region of transverse magnetic field where it moves in a circular Arc with radius 0.180m. the magnitude of the field is

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An electron in the beam of a TV picture tube is accelerated by a PD of 2.00kv. Then it passes through a region of transverse magnetic field where it moves in a circular Arc with radius 0.180m. the magnitude of the field is

An electron in the beam of a TV picture tube is accelerated by a PD of 2.00kv. Then it passes through a region of transverse magnetic field where it moves in a circular Arc with radius 0.180m. the magnitude of the field is

1 Answers

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Other Related Questions on Magnetism

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