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Find the Lorentz force of a charge 2.5C having an electric field of 5 units and magnetic field of 7.25 units with a velocity 1.5m/s.
- a)39.68
- b)68.39
- c)86.93
- d)93.68
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Find the Lorentz force of a charge 2.5C having an electric field of 5 ...
Answer: a
Explanation: The Lorentz force is given by F = qE + q(v x B), it is the sum of electric and magnetic force. On substituting q = 2.5, E = 5, v = 1.5 and B = 7.25, F = 2.5(5) + 2.5(1.5 x 7.25) = 39.68 units.
Explanation: The Lorentz force is given by F = qE + q(v x B), it is the sum of electric and magnetic force. On substituting q = 2.5, E = 5, v = 1.5 and B = 7.25, F = 2.5(5) + 2.5(1.5 x 7.25) = 39.68 units.
Most Upvoted Answer
Find the Lorentz force of a charge 2.5C having an electric field of 5 ...
Answer: a
Explanation: The Lorentz force is given by F = qE + q(v x B), it is the sum of electric and magnetic force. On substituting q = 2.5, E = 5, v = 1.5 and B = 7.25, F = 2.5(5) + 2.5(1.5 x 7.25) = 39.68 units.
Explanation: The Lorentz force is given by F = qE + q(v x B), it is the sum of electric and magnetic force. On substituting q = 2.5, E = 5, v = 1.5 and B = 7.25, F = 2.5(5) + 2.5(1.5 x 7.25) = 39.68 units.
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Community Answer
Find the Lorentz force of a charge 2.5C having an electric field of 5 ...
Lorentz force, denoted by F, is the force experienced by a charged particle when it is in the presence of both an electric field and a magnetic field. It is given by the equation:
F = q(E + v x B)
where:
- q is the charge of the particle (in coulombs),
- E is the electric field (in units),
- v is the velocity of the particle (in meters per second), and
- B is the magnetic field (in units).
In this case, we are given:
- q = 2.5 C,
- E = 5 units,
- v = 1.5 m/s, and
- B = 7.25 units.
Let's calculate the Lorentz force using the given values:
1. Calculate the cross product of v and B:
v x B = |v| |B| sin(theta) n
where:
- |v| and |B| are the magnitudes of v and B, respectively,
- theta is the angle between v and B, and
- n is the unit vector perpendicular to both v and B.
Since the angle between v and B is not given, let's assume it is 90 degrees (perpendicular). In this case, sin(theta) = 1 and n = 1.
v x B = |v| |B| n = 1.5 m/s * 7.25 units * 1 = 10.875 units
2. Calculate the Lorentz force:
F = q(E + v x B) = 2.5 C * (5 units + 10.875 units) = 2.5 C * 15.875 units = 39.6875 C*units
Rounding off to two decimal places, the Lorentz force is approximately 39.68 C*units.
Therefore, the correct answer is option A) 39.68.
F = q(E + v x B)
where:
- q is the charge of the particle (in coulombs),
- E is the electric field (in units),
- v is the velocity of the particle (in meters per second), and
- B is the magnetic field (in units).
In this case, we are given:
- q = 2.5 C,
- E = 5 units,
- v = 1.5 m/s, and
- B = 7.25 units.
Let's calculate the Lorentz force using the given values:
1. Calculate the cross product of v and B:
v x B = |v| |B| sin(theta) n
where:
- |v| and |B| are the magnitudes of v and B, respectively,
- theta is the angle between v and B, and
- n is the unit vector perpendicular to both v and B.
Since the angle between v and B is not given, let's assume it is 90 degrees (perpendicular). In this case, sin(theta) = 1 and n = 1.
v x B = |v| |B| n = 1.5 m/s * 7.25 units * 1 = 10.875 units
2. Calculate the Lorentz force:
F = q(E + v x B) = 2.5 C * (5 units + 10.875 units) = 2.5 C * 15.875 units = 39.6875 C*units
Rounding off to two decimal places, the Lorentz force is approximately 39.68 C*units.
Therefore, the correct answer is option A) 39.68.
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Find the Lorentz force of a charge 2.5C having an electric field of 5 units and magnetic field of 7.25 units with a velocity 1.5m/s.a)39.68b)68.39c)86.93d)93.68Correct answer is option 'A'. Can you explain this answer?
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