A copper rod of mass m rests on two horizontal rails distance L apart and carries a current of I from one rail to the other. The coefficient of static friction between rod and rails is `mu_(s)` What are the (a) magnitude and (b) angle (relative to the vertical) of the smallest magnetic field that puts the rod on the verge of sliding?

Answer

To solve the problem step by step, we will analyze the forces acting on the copper rod and derive the necessary equations to find the magnitude and angle of the magnetic field required to put the rod on the verge of sliding.

### Step 1: Understand the Forces Acting on the Rod
The copper rod is resting on two horizontal rails and carries a current \( I \). The forces acting on the rod include:
- The gravitational force \( mg \) acting downward.
- The magnetic force due to the current in the rod, which can be resolved into horizontal and vertical components based on the angle \( \theta \) of the magnetic field \( B \).

### Step 2: Identify the Magnetic Force Components
The magnetic force \( F_B \) acting on the rod can be calculated using the formula:
\[ F_B = I \cdot B \cdot L \]
Where:
- \( I \) is the current,
- \( B \) is the magnetic field strength,
- \( L \) is the length of the rod (distance between the rails).

The force can be resolved into two components:
- Horizontal component: \( F_{B,x} = IBL \sin(\theta) \)
- Vertical component: \( F_{B,y} = IBL \cos(\theta) \)

### Step 3: Write the Equilibrium Conditions
For the rod to be on the verge of sliding, the horizontal magnetic force must equal the maximum static friction force:
\[ F_{B,x} = \mu_s N \]
Where \( N \) is the normal force acting on the rod.

### Step 4: Determine the Normal Force
The normal force \( N \) is the sum of the gravitational force and the vertical magnetic force:
\[ N = mg + F_{B,y} = mg + IBL \cos(\theta) \]

### Step 5: Substitute the Normal Force into the Friction Equation
Substituting \( N \) into the friction equation gives:
\[ IBL \sin(\theta) = \mu_s (mg + IBL \cos(\theta)) \]

### Step 6: Rearrange the Equation
Rearranging the equation leads to:
\[ IBL \sin(\theta) = \mu_s mg + \mu_s IBL \cos(\theta) \]
\[ IBL \sin(\theta) - \mu_s IBL \cos(\theta) = \mu_s mg \]
Factoring out \( IBL \) gives:
\[ IBL (\sin(\theta) - \mu_s \cos(\theta)) = \mu_s mg \]

### Step 7: Solve for the Magnetic Field \( B \)
From the above equation, we can express \( B \):
\[ B = \frac{\mu_s mg}{IL (\sin(\theta) - \mu_s \cos(\theta))} \]

### Step 8: Find the Angle \( \theta \)
To minimize \( B \), we need to maximize the denominator \( \sin(\theta) - \mu_s \cos(\theta) \). Taking the derivative and setting it to zero gives:
\[ \frac{d}{d\theta}(\sin(\theta) - \mu_s \cos(\theta)) = 0 \]
This leads to:
\[ \cos(\theta) + \mu_s \sin(\theta) = 0 \]
Thus, we find:
\[ \tan(\theta) = -\mu_s \]
Since we are interested in the magnitude, we have:
\[ \tan(\theta) = \mu_s \]

### Step 9: Substitute \( \theta \) Back to Find \( B \)
Substituting \( \tan(\theta) = \mu_s \) into the equation for \( B \):
\[ B = \frac{\mu_s mg}{IL \left(\frac{\mu_s}{\sqrt{1+\mu_s^2}} - \mu_s \frac{1}{\sqrt{1+\mu_s^2}}\right)} \]
This simplifies to:
\[ B = \frac{\mu_s mg \sqrt{1+\mu_s^2}}{IL} \]

### Final Answers
(a) The magnitude of the magnetic field \( B \):
\[ B = \frac{\mu_s mg \sqrt{1+\mu_s^2}}{IL} \]

(b) The angle \( \theta \) relative to the vertical:
\[ \theta = \tan^{-1}(\mu_s) \]