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प्रश्न
Consider a wire carrying a steady current, I placed in a uniform magnetic field B perpendicular to its length. Consider the charges inside the wire. It is known that magnetic forces do no work. This implies that ______.
- motion of charges inside the conductor is unaffected by B since they do not absorb energy.
- some charges inside the wire move to the surface as a result of B.
- if the wire moves under the influence of B, no work is done by the force.
- if the wire moves under the influence of B, no work is done by the magnetic force on the ions, assumed fixed within the wire.
विकल्प
b and c
a and d
b and d
c and d
उत्तर
b and d
Explanation:
If a current carrying straight conductor (length l) is placed in a uniform magnetic field (B) such that it makes an angle θ with the direction of field, then the force experienced by it is Fmax= Bil sin θ. Direction of this force is obtained by the right-hand palm rule.
Right-hand palm rule: Stretch the fingers and thumb of the right hand at right angles to each other. Then if the fingers point in the direction of field B and thumb in the direction of current z, then normal to the palm will point in the direction of force
If conductor is placed perpendicular to magnetic field, then θ = 90°, Fmax = Bil
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In the product
`overset(->)("F") = "q"(overset(->)(υ) xx overset(->)("B"))`
= `"q"overset(->)(υ) xx ("B"overset(^)("i") + "B" overset(^)("j") + "B"_0overset(^)("k"))`
For q = 1 and `overset(->)(υ) = 2overset(^)("i") + 4overset(^)("j") + 6overset(^)("k")` and
`overset(->)("F") = 4overset(^)("i") - 20overset(^)("j") + 12overset(^)("k")`
What will be the complete expression for `overset(->)("B")`?
