JEE Main · 2024 · 08 Apr 2024 · Shift 1 · Magnetic Effects of Current

An electron with kinetic energy 5 eV enters a region of uniform magnetic field of

3 μ T

perpendicular to its direction. An electric field E is applied perpendicular to the direction of velocity and magnetic field. The value of E, so that electron moves along the same path, is ____

NC^{- 1}

.
(Given, mass of electron

= 9 \times 1 0^{- 31} kg

, electric charge

= 1.6 \times 1 0^{- 19} C

)

Answer: 4

Solution

For undeflected motion the electric and magnetic forces balance:

q E = q v B \Rightarrow E = v B

v = \frac{2 K E}{m} = \frac{2 ( 5 ) ( 1.6 \times 1 0 ^{- 19} )}{9 \times 1 0 ^{- 31}} = \frac{4}{3} \times 1 0^{6} m/s

E = v B = \frac{4}{3} \times 1 0^{6} \times 3 \times 1 0^{- 6} = 4 N/C

JEE Main · 2024 · 08 Apr 2024 · Shift 1 · Integer answer

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