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Q. 76

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Physics for Scientists and Engineers: A Strategic Approach with Modern Physics
Found in: Page 835

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Short Answer

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

b. What deflects a 55 g/m wire to a 12° angle when the current is 10 A?

(a)

(b)

See the step by step solution

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 , the deflection angle is 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

Here, is the mass and is the length of the wire.

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

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

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

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

Here, is the magnetic force, is the current through the wire and is the magnetic field.

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

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

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 and the direction of the magnetic field is downward.

Part (b): Step 1. Given information

Given that the mass density of the wire is , the deflection angle is and the current is .

Part (b): Step 2. Calculation

Substitute for , for , for and for into equation (6) to calculate the required magnetic field.

Part (b): Step 3. Final answer

The required magnetic field is .

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