Question

a. In FIGURE P29.76, a long, straight, current-carrying wire of linear mass density $\mu$is suspended by threads. A magnetic field perpendicular to the wire exerts a horizontal force that deflects the wire to an equilibrium angle $\theta$. Find an expression for the strength and direction of the magnetic field $\stackrel{\rightharpoonup }{B}$.

b. What $\stackrel{\rightharpoonup }{B}$deflects a 55 g/m wire to a 12° angle when the current is 10 A?

Short answer

(a) $B=\frac{\mu g\tan \theta }{I}$

(b)$0.01\text{T}$

Step-by-step solution

Part (a): Step 1. Given information

Give the diagram is as follows:

Given that the mass density of the wire is$\mu$, the deflection angle is$\theta$and a magnetic field perpendicular to the wire is applied.

Part (a): Step 2. Calculation

The formula to calculate the mass of the wire is given by

$m=\mu L...........................\left(1\right)$

Here, $m$is the mass and $L$is the length of the wire.

The formula to calculate the gravitational force of the wire is given by

$F=mg..........................\left(2\right)$

Here, $F$is the gravitational force and $g$is the acceleration due to gravity.

Substitute the expression for the mass from equation (1) into equation (2) to obtain the gravitational force.

$F=\mu Lg....................\left(3\right)$

The formula to calculate the magnetic force exerted on the wire is given by

$F_B=ILB....................\left(4\right)$

Here, $F_B$is the magnetic force, $I$is the current through the wire and $B$is the magnetic field.

The formula to calculate the deflection angle of the wire is given by

$\tan \theta =\frac{F_B}{F}........................\left(5\right)$

Substitute the expressions for $F$and $F_B$from equations (3) and (4) respectively into equation (5) and simplify to obtain the expression for the magnetic field.

$\begin{array}{rcl}\tan \theta & =& \frac{ILB}{\mu Lg}\\ B& =& \frac{\mu g\tan \theta }{I}..........................\left(6\right)\end{array}$

Also, using the right-hand rule, it can be said that the direction of the magnetic field is downward.

Part (a): Step 3. Final answer

The required magnetic field is given by$B=\frac{\mu g\tan \theta }{I}$and the direction of the magnetic field is downward.

Part (b): Step 1. Given information

Given that the mass density of the wire is $55\text{g/m}$, the deflection angle is$12°$and the current is$10\text{A}$.

Part (b): Step 2. Calculation

Substitute $55\text{g/m}$for $\mu$,$9.80{\text{m/s}}^2$for $g$, $12°$for $\theta$and $10\text{A}$for $I$into equation (6) to calculate the required magnetic field.

$\begin{array}{rcl}B& =& \frac{55\text{g/m}\times {\displaystyle \frac{1\text{kg}}{1000\text{g}}}\times 9.80{\text{m/s}}^2\times \tan 12°}{10\text{A}}\\ & \approx & 0.01\text{T}\end{array}$

Part (b): Step 3. Final answer

The required magnetic field is$0.01\text{T}$.