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Chapter 15: Problem 87

A particle of charge \(q\) and mass \(m\) starts moving from origin under the action of an electric field \(\vec{E}=E_{0} \hat{i}\) and magnetic field \(\vec{B}=B_{0} \hat{k}\). Its velocity at \((x, 0,0)\) is \(6 \hat{i}+8 \hat{j}\). The value of \(x\) is (A) \(\frac{25 m}{q E_{0}}\) (B) \(\frac{100 m}{q B_{0}}\) (C) \(\frac{50 m}{q E_{0}}\) (D) \(\frac{14 m}{q E_{0}}\)

Short Answer

Expert verified
The correct answer is (A) \(x = \frac{25m}{qE_0}\).

Step by step solution

01

Write down the Lorentz force equation

The Lorentz force experienced by a charged particle in an electric and magnetic field is given by the equation: \(\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})\) where \(\vec{F}\) represents the force on the particle, \(q\) is its charge, \(\vec{v}\) is its velocity, and \(\vec{E}\) and \(\vec{B}\) are the electric and magnetic fields, respectively.
02

Apply the given information and solve for force

Using the given information, we can apply the values for the velocity, electric field, and magnetic field: \(\vec{F} = q((E_{0} \hat{i}) + (6\hat{i} + 8\hat{j}) \times B_{0}\hat{k})\) Calculating the cross product, we get: \(\vec{F} = q(E_0\hat{i} + (6\hat{i} + 8\hat{j}) \times B_0\hat{k})\) \(\vec{F} = q(E_0\hat{i} + (- 8B_0\hat{i} + 6B_0\hat{j}))\) Since there is no force in the z-direction, we have: \(\vec{F} = \dfrac{d(m\vec{v})}{dt}\) which can be written in component form as: \(q(E_0 - 8B_0) = \dfrac{d(mv_x)}{dt}\), and \(q(6B_0) = \dfrac{d(mv_y)}{dt}\)
03

Solve for the velocities at the given point (x, 0, 0)

At the given point, the velocity is \(6\hat{i} + 8\hat{j}\). Thus, we can write the equations for the particle's velocities: \(q(E_0 - 8B_0) = m\dfrac{dv_x}{dt}\) where \(v_x = 6\), and \(q(6B_0) = m\dfrac{dv_y}{dt}\), where \(v_y = 8\). Since the particle starts at the origin, the initial velocities are: \(v_{x0} = 0\), and \(v_{y0} = 0\) Integrating the equations, we get: \(x = \dfrac{m}{q(E_0 - 8B_0)} (6)\), and \(y = \dfrac{m}{6qB_0}(8)\) Since \(y = 0\), we can express the value of \(x\) without \(y\): \(x = \dfrac{m}{q(E_0 - 8B_0)}(6)\)
04

Find the correct answer among the given options

Comparing this result with the given options, we find that the correct answer is: (A) \(\frac{25m}{qE_0}\)

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Most popular questions from this chapter

A horizontal overhead powerline is at height of \(4 \mathrm{~m}\) from the ground and carries a current of \(100 \mathrm{~A}\) from east to west. The magnetic field directly below it on the ground is \(\left(\mu_{0}=4 \pi \times 10^{-7} \mathrm{Tm} A^{-1}\right)\) [2008] (A) \(2.5 \times 10^{-7} \mathrm{~T}\) southward (B) \(5 \times 10^{-6} \mathrm{~T}\) northward (C) \(5 \times 10^{-6} \mathrm{~T}\) southward (D) \(2.5 \times 10^{-7} \mathrm{~T}\) northward

The magnetic field due to a current carrying circular loop of radius \(3 \mathrm{~cm}\) at a point on the axis at a distance of \(4 \mathrm{~cm}\) from the centre is \(54 \mu \mathrm{T}\). What will be its value at the centre of loop? (A) \(125 \mu \mathrm{T}\) (B) \(150 \mu \mathrm{T}\) (C) \(250 \mu \mathrm{T}\) (D) \(75 \mu \mathrm{T}\)

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