# An electron is moving vertically upwards with a speed of 2.0*10^8 ms^-1,magnetic field is 0.50 N A^-1 m^-1.Calculate the acceleration of the electron

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by vivek2001

plzzzzz someone help its urgent i need it

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by vivek2001

plzzzzz someone help its urgent i need it

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Magnetic force F exerted by a magnetic field *E* on a moving charge q with a velocity** v** is given by

*F* = q *v Χ B* , cross product of velocity with magnetic field.

Magnitude of the force is given by

F = q v B Sin Φ, where Φ is the angle between vectors v and B.

You have not specified the angle between*v* and** B**. Let us say vertical direction towards the sky is the Z-axis.

1. Let B be horizontal, and along positive X-axis. Then

F = 1.6 * 10⁻¹⁹ * 2 * 10⁸ * 0.50 Sin 90° = = 1.6 * 10⁻¹¹ N

Force will be along positive Y-axis, as given by the right hand thumb rule.

Acceleration = F/mass = 1.6 * 10⁻¹¹/9.11*10⁻³¹ = 1.756 * 10¹⁹ m/sec²

2. Let B be horizontal , and along positive Y-axis , then

F = 1.6 * 10⁻¹⁹ * 2 * 10⁸ * 0.50 Sin 90° = 1.6 * 10⁻¹¹ N

Force will be along negative X-axis, as given by the right hand thumb rule.

Acceleration = F/mass = 1.6 * 10⁻¹¹/9.11*10⁻³¹ = 1.756 * 10¹⁹ m/sec²

3. Let B be along vertical direction, then Ф = 0°. Hence

F = 1.6 * 10⁻¹⁹ * 2 * 10⁸ * 0.50 Sin 0° = 0

Magnitude of the force is given by

F = q v B Sin Φ, where Φ is the angle between vectors v and B.

You have not specified the angle between

1. Let B be horizontal, and along positive X-axis. Then

F = 1.6 * 10⁻¹⁹ * 2 * 10⁸ * 0.50 Sin 90° = = 1.6 * 10⁻¹¹ N

Force will be along positive Y-axis, as given by the right hand thumb rule.

Acceleration = F/mass = 1.6 * 10⁻¹¹/9.11*10⁻³¹ = 1.756 * 10¹⁹ m/sec²

2. Let B be horizontal , and along positive Y-axis , then

F = 1.6 * 10⁻¹⁹ * 2 * 10⁸ * 0.50 Sin 90° = 1.6 * 10⁻¹¹ N

Force will be along negative X-axis, as given by the right hand thumb rule.

Acceleration = F/mass = 1.6 * 10⁻¹¹/9.11*10⁻³¹ = 1.756 * 10¹⁹ m/sec²

3. Let B be along vertical direction, then Ф = 0°. Hence

F = 1.6 * 10⁻¹⁹ * 2 * 10⁸ * 0.50 Sin 0° = 0