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Blizzard [7]
2 years ago
8

A wire of length 6cm makes an angle of 20° with a 3 mT

Physics
1 answer:
Crazy boy [7]2 years ago
5 0

Answer:

Approximately 7.3 \times 10^{-3}\; \rm A (approximately 7.3\; \rm mA) assuming that the magnetic field and the wire are both horizontal.

Explanation:

Let \theta denote the angle between the wire and the magnetic field.

Let B denote the magnitude of the magnetic field.

Let l denote the length of the wire.

Let I denote the current in this wire.

The magnetic force on the wire would be:

F = B \cdot l \cdot I \cdot \sin(\theta).

Because of the \sin(\theta) term, the magnetic force on the wire is maximized when the wire is perpendicular to the magnetic field (such that the angle between them is 90^\circ.)

In this question:

  • \theta = 20^\circ (or, equivalently, (\pi / 9) radians, if the calculator is in radian mode.)
  • B = 3\; \rm mT = 3 \times 10^{-3}\; \rm T.
  • l = 6\; \rm cm = 6 \times 10^{-2}\;\rm m.
  • F = 1.5\times 10^{-4}\; \rm N.

Rearrange the equation F = l \cdot I \cdot \sin(\theta) to find an expression for I, the current in this wire.

\begin{aligned} I &= \frac{F}{l \cdot \sin(\theta)} \\ &= \frac{3\times 10^{-3}\; \rm T}{6 \times 10^{-2}\; \rm m \times \sin \left(20^{\circ}\right)} \\ &\approx 7.3 \times 10^{-3}\; \rm A = 7.3 \; \rm mA\end{aligned}.

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