Question

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

magnetic field. What current is needed to cause an
upward force of 1.5 x 10-4 N?

Answer 1

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

Step-by-step 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 * 10^(-3)\; \rm T`.
• 
  `l = 6\; \rm cm = 6 * 10^(-2)\;\rm m`.
• 
  `F = 1.5* 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 &= (F)/(l \cdot \sin(\theta)) \\ &= (3* 10^(-3)\; \rm T)/(6 * 10^(-2)\; \rm m * \sin \left(20^(\circ)\right)) \\ &\approx 7.3 * 10^(-3)\; \rm A = 7.3 \; \rm mA\end{aligned}`.