Question

Find the radius of trajectory of a proton moving with velocity \(4\times10^{5}\,m\,s^{-1}\) in a magnetic field of \(0.01\,T\).

• A. \(0.2\,m\)
• B. \(0.8\,m\)
• C. \(0.6\,m\)
• D. \(0.4\,m\)

Hint:

Remember the circular motion formula in a magnetic field: \[ r=\frac{mv}{qB}. \] Larger velocity increases radius, while stronger magnetic field decreases radius.

Answer 1

The Correct Option is D

Concept: A charged particle moving perpendicular to a magnetic field follows a circular path. The radius of the path is given by \[ r=\frac{mv}{qB} \] where \(m\) is mass, \(q\) is charge and \(B\) is magnetic field.

Step 1: Write the given values. For a proton, \[ m=1.67\times10^{-27}\,kg \] \[ q=1.6\times10^{-19}\,C \] Given, \[ v=4\times10^5\,m\,s^{-1} \] \[ B=0.01\,T \]

Step 2: Substitute into the formula. \[ r= \frac{1.67\times10^{-27}\times4\times10^5} {1.6\times10^{-19}\times0.01} \] \[ = \frac{6.68\times10^{-22}} {1.6\times10^{-21}} \] \[ =4.175\times10^{-1} \] \[ \approx0.42\,m \]

Step 3: Choose nearest option. \[ r\approx0.4\,m \] Therefore, \[ \boxed{(D)} \]