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Numerical

Derive the expression for the force on a current-carrying conductor in a magnetic field.

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#### Solution

- The force experienced is equal to the sum of Lorentz forces on the individual charge carriers in the wire.
- Consider a small length of wire ‘dl’, area – A, Current – I.
- The free electrons drift opposite to the direction of the current.
- The relation between I & V
_{d}is given by

I = neAV_{d } - In magnetic field, the force experienced is given by

`vec"F" = - "e" (vec"v"_"d" xx vec"B")`

n → number of free electrons per unit volume.

n = `"N"/"V"`

V = A dl - Lorentz force = n A dl × – `"e"(vec"v"_"d" xx vec"B")`

dF = -enA dl `(vec"v"_"d" xx vec"B")` - Current element, `"I"vec"dl" = - "enA"vec"v"_"d"`

`therefore "d"vec"F" = ("l" vec"dl" = vec"B")`

`vec"F" = ("l" vec"l" xx vec"B")`

F = BIl sin θ**Case (i):**

If the conductor is along the magnetic field; θ = 0° ; F = 0.**Case (ii):**

If the conductor is perpendicular to the magnetic field; θ = 90°; F = BIl.

Concept: Magnetic Effects of Current

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