16E. - University Physics with Modern Physics

Question

Question: An electron at point A in Fig. E27.15 has a speed \({v_0}\) of 1 41 10 m/s. Find (a) the magnitude and direction of the magnetic field that will cause the electron to follow the semicircular path from A to B, and 

(b) the time required for the electron to move from A to B.

 

Repeat the exercise 27.15 for the case in which the particle is a proton rather than an electron.

Step-by-Step Solution

Answer

(a) The magnitude of the magnetic field is \(0.29\;{\rm{T}}\) and is directed out of the page.

Step 1: Given data

Speed of the proton

\({v_0} = 1.41 \times {10^6}\;{{\rm{m}}\mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right./{\rm{s}}}\) 

The radius of the semi-circular path\(\begin{aligned}{c}r = 5\;{\rm{cm}}\\ = 5 \cdot \left( {1\;{\rm{cm}} \times \frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right)\\ = 0.05\;{\rm{m}}\end{aligned}\) 

Step 2: Magnitude of magnetic field that causes a particle to move in a circular path

The magnitude of the magnetic field that causes a charge \(q\) of mass \(m\) and velocity \(v\) to move in a circular path of radius \(r\) is

\(B = \frac{{mv}}{{qr}}\)                                                                                                        .....(i)

Step 3: (a) Determination of the magnitude and direction of magnetic field

The charge of the proton is

\(q = 1.6 \times {10^{ - 19}}\;{\rm{C}}\) 

The mass of the proton is

\(m = 1.67 \times {10^{ - 27}}\;{\rm{kg}}\) 

From equation (i), the magnitude of the magnetic field is

\(\begin{aligned}B = \frac{{1.67 \times {{10}^{ - 27}}\;{\rm{kg}} \times 1.41 \times {{10}^6}\;{{\rm{m}}\mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.{\rm{s}}}}}{{1.6 \times {{10}^{ - 19}}\;{\rm{C}} \times 0.05\;{\rm{m}}}}\\ = 0.29 \cdot \left( {1\;{{{\rm{kg}}} \mathord{\left/{\vphantom {{{\rm{kg}}} {{\rm{C}}\cdot }}} \right.{{\rm{C}} \cdot }}{\rm{s}} \times \frac{{1\;{\rm{T}}}}{{1\;{{{\rm{kg}}} \mathord{\left/{\vphantom {{{\rm{kg}}} {{\rm{C}} \cdot }}} \right.{{\rm{C}} \cdot }}{\rm{s}}}}} \right)\\ =0.29{\rm{T}}\end{aligned}\) 

Thus, the required magnitude is \(0.29\;{\rm{T}}\).

From the diagram it is evident that the initial force is towards the right. The direction of magnetic force on a positive charge is the direction of cross product of the velocity to the magnetic field. Thus, it is evident that the magnetic field is directed out of the page.