A straight copper wire of length 3.0 cm carries a current of 5 A and is placed in a uniform magnetic field of strength 0.40 T. The wire is perpendicular to the field. Find the magnitude of the magnetic force acting on the wire.
Answer
0.06 N
The correct answer is 0.06 N — Hence the magnetic force is (0.06) N.
Step-by-step solution
- Convert the length from cm to m: 3.0 cm = 0.03 m.
- Use the magnetic force formula for a straight conductor: (F=IBL).
- Insert the values: (I=5) A, (B=0.40) T, (L=0.03) m.
- Calculate (F = 5 \times 0.40 \times 0.03 = 0.06) N.
- Hence the magnetic force is (0.06) N.
Common mistake: Forgetting to convert the length from centimetres to metres before plugging into the formula.
Concept tested: Magnetic force on a current‑carrying conductor
Published · How this solution was prepared and checked
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