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Find the magnetic force when a charge 3.5C with flux density of 4 units is having a velocity of 2m/s.
- a)14
- b)28
- c)7
- d)32
Correct answer is option 'B'. Can you explain this answer?
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Find the magnetic force when a charge 3.5C with flux density of 4 unit...
The magnetic force is given by F = q(v x B), where q = 3.5C, v = 2m/s and B = 4 units. Thus we get F = 3.5(2 x 4) = 28 units.
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Find the magnetic force when a charge 3.5C with flux density of 4 unit...
Magnetic Force on a Charged Particle
The magnetic force on a charged particle moving through a magnetic field can be calculated using the formula:
F = q * (v x B)
Where:
F is the magnetic force
q is the charge of the particle
v is the velocity of the particle
B is the magnetic field or flux density
Given:
Charge (q) = 3.5 C
Flux density (B) = 4 units
Velocity (v) = 2 m/s
Calculating the Magnetic Force:
To calculate the magnetic force, we need to determine the cross product of the velocity and the magnetic field vectors.
Step 1: Convert the given values into vector form.
q = 3.5 C
v = 2 m/s (velocity vector)
B = 4 units (magnetic field vector)
Step 2: Determine the cross product of the velocity and magnetic field vectors.
v x B = |v| |B| sin(θ) n
Where:
|v| is the magnitude of the velocity vector
|B| is the magnitude of the magnetic field vector
θ is the angle between the velocity and magnetic field vectors
n is the unit vector perpendicular to both v and B
Since the angle between v and B is not given, we assume it to be 90°, which means sin(θ) = 1.
v x B = |v| |B| n
Step 3: Calculate the magnitude of the cross product.
|v x B| = |v| |B|
Given:
|v| = 2 m/s
|B| = 4 units
|v x B| = 2 * 4 = 8 units
Step 4: Calculate the magnetic force using the formula.
F = q * |v x B|
Given:
q = 3.5 C
|v x B| = 8 units
F = 3.5 C * 8 units = 28 C units
Therefore, the magnetic force on the charged particle is 28 C units.
Answer: The correct answer is option 'B' (28).
The magnetic force on a charged particle moving through a magnetic field can be calculated using the formula:
F = q * (v x B)
Where:
F is the magnetic force
q is the charge of the particle
v is the velocity of the particle
B is the magnetic field or flux density
Given:
Charge (q) = 3.5 C
Flux density (B) = 4 units
Velocity (v) = 2 m/s
Calculating the Magnetic Force:
To calculate the magnetic force, we need to determine the cross product of the velocity and the magnetic field vectors.
Step 1: Convert the given values into vector form.
q = 3.5 C
v = 2 m/s (velocity vector)
B = 4 units (magnetic field vector)
Step 2: Determine the cross product of the velocity and magnetic field vectors.
v x B = |v| |B| sin(θ) n
Where:
|v| is the magnitude of the velocity vector
|B| is the magnitude of the magnetic field vector
θ is the angle between the velocity and magnetic field vectors
n is the unit vector perpendicular to both v and B
Since the angle between v and B is not given, we assume it to be 90°, which means sin(θ) = 1.
v x B = |v| |B| n
Step 3: Calculate the magnitude of the cross product.
|v x B| = |v| |B|
Given:
|v| = 2 m/s
|B| = 4 units
|v x B| = 2 * 4 = 8 units
Step 4: Calculate the magnetic force using the formula.
F = q * |v x B|
Given:
q = 3.5 C
|v x B| = 8 units
F = 3.5 C * 8 units = 28 C units
Therefore, the magnetic force on the charged particle is 28 C units.
Answer: The correct answer is option 'B' (28).
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Find the magnetic force when a charge 3.5C with flux density of 4 units is having a velocity of 2m/s.a)14b)28c)7d)32Correct answer is option 'B'. Can you explain this answer?
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