Magnetic Effects of Current — CBSE Class 12 · 2024 Board Question with Solution

An electron moving with a velocity

v = (1.0 \times 1 0^{7} m/s) \hat{i} + (0.5 \times 1 0^{7} m/s) \hat{j}

enters a region of uniform magnetic field

B = (0.5 mT) \hat{j}

. Find the radius of the circular path described by it. While rotating, does the electron trace a linear path too? If so, calculate the linear distance covered by it during the period of one revolution.

Solution

Only the velocity component perpendicular to

B

produces the circular motion. Since

B

is along

\hat{j}

, the perpendicular part is

v_{⊥} = 1.0 \times 1 0^{7} m/s

(the

\hat{i}

component) and the parallel part is

v_{∥} = 0.5 \times 1 0^{7} m/s

(the

\hat{j}

component).
Radius of the circular path:

r = \frac{m v _{⊥}}{q B} = \frac{( 9.1 \times 1 0 ^{- 31} ) ( 1.0 \times 1 0 ^{7} )}{( 1.6 \times 1 0 ^{- 19} ) ( 0.5 \times 1 0 ^{- 3} )} \approx 0.114 m .

The parallel component is unaffected by the field, so the electron drifts steadily along

\hat{j}

while circling - the actual path is a helix (a straight-line drift superposed on the circle).
The time for one revolution is

T = \frac{2 π m}{q B} = \frac{2 π ( 9.1 \times 1 0 ^{- 31} )}{( 1.6 \times 1 0 ^{- 19} ) ( 0.5 \times 1 0 ^{- 3} )} \approx 7.15 \times 1 0^{- 8} s .

Linear distance in one revolution (pitch):

p = v_{∥} T = (0.5 \times 1 0^{7}) (7.15 \times 1 0^{- 8}) \approx 0.36 m .

r \approx 0.114 m, p \approx 0.36 m

CBSE Class 12 Board · 2024 · Subjective

CBSE Class 12 · 2024 · Magnetic Effects of CurrentLong Answer

Solution

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