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What is the force on a 2 m long conductor carrying a current of 5 A in a magneti

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Question: What is the force on a 2 m long conductor carrying a current of 5 A in a magnetic field of 0.3 T at an angle of 30° to the field? (2021)

Options:

  1. 0.75 N
  2. 1.5 N
  3. 2.5 N
  4. 3.0 N

Correct Answer: 1.5 N

Exam Year: 2021

Solution: 

Using the formula F = BIL sin(θ), F = 0.3 * 2 * 5 * sin(30°) = 1.5 N.

What is the force on a 2 m long conductor carrying a current of 5 A in a magneti

Practice Questions

Q1
What is the force on a 2 m long conductor carrying a current of 5 A in a magnetic field of 0.3 T at an angle of 30° to the field? (2021)
  1. 0.75 N
  2. 1.5 N
  3. 2.5 N
  4. 3.0 N

Questions & Step-by-Step Solutions

What is the force on a 2 m long conductor carrying a current of 5 A in a magnetic field of 0.3 T at an angle of 30° to the field? (2021)
  • Step 1: Identify the values given in the problem. We have a length of the conductor (L) = 2 m, current (I) = 5 A, magnetic field strength (B) = 0.3 T, and the angle (θ) = 30°.
  • Step 2: Write down the formula to calculate the force (F) on the conductor in a magnetic field. The formula is F = BIL sin(θ).
  • Step 3: Substitute the known values into the formula. F = 0.3 T * 2 m * 5 A * sin(30°).
  • Step 4: Calculate sin(30°). The value of sin(30°) is 0.5.
  • Step 5: Now substitute sin(30°) into the equation: F = 0.3 * 2 * 5 * 0.5.
  • Step 6: Perform the multiplication: 0.3 * 2 = 0.6, then 0.6 * 5 = 3, and finally 3 * 0.5 = 1.5.
  • Step 7: The final answer is F = 1.5 N.
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