Unit 12: Magnetism and Electromagnetic Induction
Weightage : 15–18%
Each of the figures below shows the path of a charged particle moving in the plane of the page in a magnetic field that is perpendicular to the page. If the mass, speed, and charge of the particles are the same, in which case does the field have the greatest magnitude?
Correct Answer: \( \boxed{\mathrm{D}} \)
For a charged particle moving perpendicular to a uniform magnetic field, the magnetic force provides the centripetal force:
\( qvB=\dfrac{mv^2}{r}. \)
Solving for the magnetic field gives
\( B=\dfrac{mv}{qr}. \)
Since the mass \(m\), speed \(v\), and charge \(q\) are the same in all four cases, the magnetic field is inversely proportional to the radius of curvature:
\( B\propto\dfrac{1}{r}. \)
Therefore, the strongest magnetic field corresponds to the path with the smallest radius of curvature (the sharpest turn).
Among the four paths shown, Case D has the smallest radius of curvature, indicating the greatest magnetic force and, therefore, the greatest magnetic field.
Hence, the correct answer is (D).
An ion with charge \(q\), mass \(m\), and speed \(v\) enters a magnetic field \(B\) and is deflected into a path with a radius of curvature \(R\). If a second ion has speed \(2v\), while \(m\), \(q\), and \(B\) are unchanged, what will be the radius of the second ion’s path?
Correct Answer: \( \boxed{\mathrm{B}} \)
The magnetic force provides the centripetal force:
\( qvB=\dfrac{mv^2}{r}. \)
Solving for the radius gives
\( r=\dfrac{mv}{qB}. \)
Since \(m\), \(q\), and \(B\) remain constant, the radius is directly proportional to the particle’s speed:
\( r\propto v. \)
If the second ion has speed
\( v’=2v, \)
then its radius becomes
\( r’=\dfrac{m(2v)}{qB}=2r=2R. \)
Therefore, doubling the speed doubles the radius of the circular path.
Hence, the correct answer is (B).
An electron moves in the plane of the page through two regions of space along the dotted-line trajectory shown in the figure. There is a uniform electric field in Region I directed into the plane of the page (as shown). There is no electric field in Region II. What is a necessary direction of the magnetic field in Regions I and II? Ignore gravitational forces.
| Option | Region I | Region II |
|---|---|---|
| (A) | Toward bottom of page | Up on the page |
| (B) | Toward top of page | Into the page |
| (C) | Toward top of page | Out of the page |
| (D) | Toward bottom of page | Out of the page |
Correct Answer: \( \boxed{\mathrm{C}} \)
Region I:
The electric field is directed into the page. Since the particle is an electron (\(q<0\)), the electric force is opposite the electric field, i.e., out of the page.
The electron travels straight through Region I, so the magnetic force must exactly oppose the electric force. Therefore, the magnetic force must be into the page.
Using the magnetic force equation
\( \vec{F}=q\vec{v}\times\vec{B}, \)
with the electron moving to the right, the magnetic field must point toward the top of the page so that the magnetic force on the electron is into the page.
Region II:
There is no electric field, so the only force is magnetic. The electron initially moves to the right and curves upward, meaning the magnetic force is upward.
Applying \( \vec{F}=q\vec{v}\times\vec{B} \) (or the left-hand rule for an electron), a magnetic field out of the page produces an upward force on the electron.
Therefore:
Region I: Toward the top of the page
Region II: Out of the page
Hence, the correct answer is (C).
