Question

A uniform magnetic field of 1.5 T exists in a cylindrical region of radius 10.0 cm, its direction parallel to the axis along east to west. A wire carrying current of 7.0 A in the north to south direction passes through this region. What is the magnitude and direction of the force on the wire if,

(a) the wire intersects the axis,

(b) the wire is turned from N-S to northeast-northwest direction,

(c) the wire in the N-S direction is lowered from the axis by a distance of 6.0 cm?

• A. (a) Magnitude: 2.1 N, Direction: Down; (b) Magnitude: 1.485 N, Direction: Up; (c) Magnitude: 1.68 N, Direction: Down
• B. (a) Magnitude: 2.1 N, Direction: Up; (b) Magnitude: 1.485 N, Direction: Down; (c) Magnitude: 1.68 N, Direction: Down
• C. (a) Magnitude: 2.1 N, Direction: Down; (b) Magnitude: 1.485 N, Direction: Down; (c) Magnitude: 1.68 N, Direction: Up
• D. (a) Magnitude: 1.485 N, Direction: Down; (b) Magnitude: 2.1 N, Direction: Up; (c) Magnitude: 1.68 N, Direction: Down

Answer 1

Answer: (A)

(a) Magnitude: 2.1 N, Direction: Down; (b) Magnitude: $\frac{2.1}{\sqrt{2}}$ N (approximately 1.485 N), Direction: Up; (c) Magnitude: 1.68 N, Direction: Down

Answer 2

The magnetic force on a current-carrying wire is given by $\vec{F} = I(\vec{L} \times \vec{B})$.

Coordinate System and Directions:

• East: +x-axis
• West: -x-axis
• North: +y-axis
• South: -y-axis
• Up: +z-axis
• Down: -z-axis

Given Values:

• Magnetic field strength, $B = 1.5$ T
• Radius of the cylindrical region, $R = 10.0$ cm $= 0.10$ m
• Current in the wire, $I = 7.0$ A

Problem Setup: The magnetic field is along the East to West direction, parallel to the axis of the cylinder. Thus, $\vec{B} = -1.5 \hat{i}$ T. The magnetic field exists in the region where $y^2 + z^2 \le R^2$.

(a) The wire intersects the axis: The wire is along the North-South direction (y-axis) and passes through the origin. The length of the wire inside the magnetic field is $L = 2R = 2 \times 0.10 \text{ m} = 0.20$ m. The current is from North to South, so $\vec{L} = L (-\hat{j}) = (0.20 \text{ m}) (-\hat{j})$. $\vec{F} = I (\vec{L} \times \vec{B}) = (7.0 \text{ A}) \times [(0.20 \text{ m}) (-\hat{j}) \times (1.5 \text{ T}) (-\hat{i})]$ $\vec{F} = (7.0 \times 0.20 \times 1.5) \times [(-\hat{j}) \times (-\hat{i})] \text{ N} = 2.1 \times (\hat{k}) \text{ N}$ Wait, $(-\hat{j}) \times (-\hat{i}) = \hat{j} \times \hat{i} = -\hat{k}$. $\vec{F} = 2.1 \times (-\hat{k}) \text{ N} = -2.1 \hat{k} \text{ N}$ Magnitude: $2.1$ N, Direction: $-\hat{k}$ (Down).

(b) The wire is turned from N-S to northeast-northwest direction: This implies the wire is along the Northeast-Southwest line. Assuming current flows from Southwest to Northeast, $\vec{L}$ is in the direction $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$. The length inside the field is $L = 2R = 0.20$ m. $\vec{L} = L \left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right) = (0.20 \text{ m}) \left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right)$ $\vec{F} = I (\vec{L} \times \vec{B}) = (7.0 \text{ A}) \times \left[(0.20 \text{ m}) \left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right) \times (1.5 \text{ T}) (-\hat{i})\right]$ $\vec{F} = \frac{7.0 \times 0.20 \times 1.5}{\sqrt{2}} \times [(\hat{i}+\hat{j}) \times (-\hat{i})] \text{ N} = \frac{2.1}{\sqrt{2}} \times [\hat{i}\times(-\hat{i}) + \hat{j}\times(-\hat{i})] \text{ N}$ $\vec{F} = \frac{2.1}{\sqrt{2}} \times [0 - (-\hat{k})] \text{ N} = \frac{2.1}{\sqrt{2}} \hat{k} \text{ N}$ Magnitude: $\frac{2.1}{\sqrt{2}} \approx 1.485$ N, Direction: $+\hat{k}$ (Up).

(c) The wire in the N-S direction is lowered from the axis by a distance of 6.0 cm: The wire is parallel to the y-axis, but at $z = -0.06$ m. The wire is within the field if $y^2 + (-0.06)^2 \le (0.10)^2$, which means $y^2 \le 0.0100 - 0.0036 = 0.0064$. So, $-0.08 \le y \le 0.08$. The length of the wire inside the field is $L = 0.08 - (-0.08) = 0.16$ m. The current is from North to South, so $\vec{L} = L (-\hat{j}) = (0.16 \text{ m}) (-\hat{j})$. $\vec{F} = I (\vec{L} \times \vec{B}) = (7.0 \text{ A}) \times [(0.16 \text{ m}) (-\hat{j}) \times (1.5 \text{ T}) (-\hat{i})]$ $\vec{F} = (7.0 \times 0.16 \times 1.5) \times [(-\hat{j}) \times (-\hat{i})] \text{ N} = 1.68 \times (-\hat{k}) \text{ N}$ Magnitude: $1.68$ N, Direction: $-\hat{k}$ (Down).

Summary: (a) Magnitude: 2.1 N, Direction: Down (b) Magnitude: $\frac{2.1}{\sqrt{2}}$ N $\approx 1.485$ N, Direction: Up (c) Magnitude: 1.68 N, Direction: Down