Question

The force on a wire is a maximum of 5.50×10^{−2} NNwhen
placed between the pole faces of a magnet. The current flows
horizontally to the right and the magnetic field is vertical. The
wire is observed to "jump" toward the observer when the current is
turned on.

If the wire is tipped so that it makes an angle of 14.0 ∘∘with
the horizontal, what force will it now feel? [*Hint*: What
length of wire will now be in the field?]

Force( rotated) = ?

Express your answer to three significant figures and include the appropriate units.

Answer #1

**Ans:****0.0534
N**

**Explanation:**

We know, for a current carrying wire of length L , carrying a current of I in horizontal direction in a magnetic field of B directed in vartical direction, the magnitude of magnetic force on the wire will be , F =BIL.......(1)

As the current is flowing in horizontal direction, the wire is aligned in horizontal direction.

Now if the wire is tipped so that it makes an angle of 14° with horizontal, the component of the length of wire which is now within the field will be , L' = L×cos(14°)

So magnetic force will be now,

F' = BIL' = BLL×cos(14°) = F×cos(14°)

Here,

The force on the wise will be now 0.0534 N.

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