A wire of length 10 cm lying normal to the magnetic field of 0.5T is experiencing a force of 5N. The current in the wire is
Correct answer: C. 100A
- A. 10A
- B. 50A
- C. 100A
- D. 500A
Explanation
Given that:Length = L = 10cm = 10 × 10-2 mMagnetic field = B = 0.5TForce = F = 5NCurrent = I =?The force experienced by a current-carrying wire in a magnetic field can be calculated using the formula:F = ILBI = F / LBPutting the values in the above equation, we get:I = 5 / 10× 10-2 × 0.5I = 100 A
This question appeared on the ETEA MDCAT 2010 paper, which you can sit online with every answer explained.
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About Magnetic Flux Density
Magnetic flux density measures the magnetic force acting on a current carrying conductor or moving charge in a magnetic field. Work includes the relations B equals F divided by IL and F equals qvB, the tesla as its SI unit, field direction and the distinction between flux density and magnetic flux.
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