I need help with all of these parts, because I don't understand it cause you help me with it and can you label it **Problem 2:** An electron (mass \( m \), charge \(-e\)) in a cathode-ray tube is accelerated through a potential difference of \(\Delta V = 10 \text{ kV}\), then passes through the \( d = 2.0 \text{ cm} \) wide region of uniform magnetic field shown in Fig. 1. What field strength will deflect the electron by 10°?a) Find the expression for the speed \( v \) to which the electron is accelerated when moving between the electrodes in terms of \(\Delta V\), \( m \), and \( e \).b) When crossing the region with a magnetic field, the electron moves along a circular trajectory of radius \( r \) with the center at point \( O \), as shown in Fig.1. Express \( r \) in terms of the size of the magnetic field region \( d \) and the angle \(\theta\).c) Using the formula for the radius of the cyclotron orbit, \( r = \frac{mv}{eB} \), and your answers to the previous questions, work out a symbolic formula for \( B \) as a function of \( d \), \(\Delta V\), \(\theta\), \( m \), and \( e \).d) For the electron to be deflected by 10° with respect to its original direction, the angle subtended by the circular part of its trajectory at point \( O \) has to be equal to \(\theta = 10^\circ\) (make sure that you understand why this is so). Use this and the numerical values of other parameters to compute \( B \). (Answer: \( B = 2.9 \text{ mT} \)).**FIG. 1:** The scheme for Problem 2 features a diagram showing an electron moving through an electric potential difference and subsequently entering a region with a magnetic field, indicated with blue arrows and 'X' representing field lines going into the page. The diagram illustrates the circular path and the deflection angle \(\theta = 10^\circ\) from the original direction. The region width is \( d = 2 \text{ cm} \).

a) Since the electron is accelerated by a voltage ∆V, we have 12mv2=e∆V⇒v=2e∆Vm The is the expression for the speed of electron when it enters the magnetic field.b) From the diagram we have rsinθ=2 cm = 0.02 m⇒r=0.02sinθc) In the magnetic field , the electron moves in a circular trajectory whose radius is given by the expression, mv2r=evB⇒r=mveB=meB2e∆Vm=1B2m∆Ve⇒0.02sinθ=1B2m∆Ve⇒B=sinθ0.022m∆Ved) Entering the numerical values we have, B=sinθ0.022m∆Ve=sin100.022·9.1·10-31·10·1031.6·10-19=0.00293 Tesla This is the required field.