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Chapter 15: Problem 102

A conducting \(\operatorname{rod} A B\) of length \(l=1 \mathrm{~m}\) is moving at a velocity \(v=4 \mathrm{~m} / \mathrm{s}\) making an angle \(30^{\circ}\) with its length. \(C\) is the middle point of the rod. A uniform magnetic field \(B=2 \mathrm{~T}\) exists in a direction perpendicular to the plane of motion, then (A) \(v_{A}-v_{B}=8 \mathrm{v}\) (B) \(v_{A}-v_{B}=4 \mathrm{v}\) (C) \(v_{B}-v_{C}=2 \mathrm{v}\) (D) \(v_{B}-v_{C}=-2 \mathrm{v}\)

Short Answer

Expert verified
The correct answer is (D) \(v_{B} - v_{C} = -2v\).

Step by step solution

01

Analyze the given information

We are given the following data: Length of the rod, l = 1 m Velocity of the rod, v = 4 m/s Angle the rod makes with its length, 胃 = 30掳 Magnetic field perpendicular to the plane of motion, B = 2 T Since the magnetic field is uniform and perpendicular to the plane of motion, we will focus on finding the velocities of points A, B, and C.
02

Determine the velocity components of point A

Let's denote the velocity of point A as \(v_A\). As the rod is making a 30掳 angle with its length, we need to find the horizontal and vertical components of the velocity of point A. Horizontal component of A,\(v_{Ax} = v\cos(30^\circ) = 4\cos(30^\circ) = 4(\frac{\sqrt{3}}{2}) =2\sqrt{3} m/s\) Vertical component of A, \(v_{Ay} = v\sin(30^\circ) = 4\sin(30^\circ) = 4(\frac{1}{2}) = 2 m/s\) Thus, the velocity of point A, \(v_A = (2\sqrt{3}, 2) m/s\)
03

Determine the velocity components of point B

Let's denote the velocity of point B as \(v_B\). Notice that point B has the same velocity as point A but with opposite direction: Horizontal component of B, \(v_{Bx} = -v_{Ax} = -2\sqrt{3} m/s\) Vertical component of B, \(v_{By} = -v_{Ay} = -2 m/s\) Thus, the velocity of point B, \(v_B = (-2\sqrt{3}, -2) m/s\)
04

Determine the velocity components of point C

Let's denote the velocity of point C (the midpoint of the rod) as \(v_C\). The middle point has a horizontal component of velocity equal to half of the horizontal velocity difference of A and B. Horizontal component of C, \(v_{Cx} = \frac{v_{Ax} - v_{Bx}}{2} = \frac{(2\sqrt{3}) - (-2\sqrt{3})}{2} = 2\sqrt{3} m/s\) Since point C is at the middle of the rod, and the rod translates, the vertical component of velocity of point C will be the same as for A or B. Vertical Component of C, \(v_{Cy} = v_{Ay} = 2 m/s\) Thus, the velocity of point C, \(v_C = (2\sqrt{3}, 2) m/s\)
05

Calculate the velocity differences and compare with given options

Option (A): \(v_A - v_B = (4\sqrt{3}, 4) m/s\), but this is not equal to \(8v\) Option (B): \(v_A - v_B = (4\sqrt{3}, 4) m/s\), but this is not equal to \(4v\) Option (C): \(v_B - v_C = (-4\sqrt{3}, -4) m/s\), but this is not equal to \(2v\) Option (D): \(v_B - v_C = (-4\sqrt{3}, -4) m/s\), and this is equal to \(-2v\), since \(v = (2\sqrt{3}, 2) m/s\) So, the correct answer is (D).

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