Question 5.58: In Figure 5.98, a straight current-carrying conductor lies a...
In Figure 5.98, a straight current-carrying conductor lies at right-angles to a magnetic field of flux density 1.2 T such that 250 mm of its length lies within the field. If the current passing through the conductor is 15 A, determine the force on the conductor and the direction of its motion.
In order to find the magnitude of the force we use the relationship F = BIl, hence:
F = BIl = 1.2 × 15 × 250 × 10^{−3} = 4.5 N
Now the direction of motion is easily found using Fleming’s left-hand rule, where we know that the first finger points in the direction of the magnetic field north and south, the second finger points inwards into the page in the direction of the current, which leaves your thumb pointing downwards in the direction of motion.
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