As illustrated in the figure, a convex lens of focal length 12 ext cm is situated within a uniform magnetic field B of magnitude 1.2 ext T oriented parallel to its principal axis. A particle with a mass of 2.0 imes 10^

Question

Quantrex Academy · Accepted Answer

Correct answer: C. 8 ext cm

The radius of the circular path of the charged particle in the magnetic field is given by:

R = \dfrac{mv}{qB}

Substituting the given values:

R = \dfrac{2.0 \times 10^{-5} \times 4.8}{2.0 \times 10^{-3} \times 1.2} = \dfrac{9.6 \times 10^{-5}}{2.4 \times 10^{-3}} = 4 \times 10^{-2} \text{ m} = 4 \text{ cm}

The circular path of the particle acts as an object for the convex lens. The plane of this circle is perpendicular to the principal axis.
Given the object distance

u = -18 \text{ cm}

and the focal length of the convex lens

f = +12 \text{ cm}

.

Using the lens formula:

\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}

\dfrac{1}{v} - \dfrac{1}{-18} = \dfrac{1}{12}

\dfrac{1}{v} + \dfrac{1}{18} = \dfrac{1}{12}

\dfrac{1}{v} = \dfrac{1}{12} - \dfrac{1}{18} = \dfrac{3 - 2}{36} = \dfrac{1}{36}

v = 36 \text{ cm}

The transverse magnification

m

is:

m = \dfrac{v}{u} = \dfrac{36}{-18} = -2

The radius of the circle along which the image moves is:

R' = |m| \times R = |-2| \times 4 \text{ cm} = 8 \text{ cm}

As illustrated in the figure, a convex lens of focal length 12 ext...

Options

Correct answer

Step-by-step solution

More from Magnetic Field