JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Magnetic Effects of Current

A proton is moving undeflected in a region of crossed electric and magnetic fields at a constant speed of

2 \times 1 0^{5} ms^{- 1}

. When the electric field is switched off, the proton moves along a circular path of radius 2 cm . The magnitude of electric field is

x \times 1 0^{4} N / C

. the value of x is ______.
Take the mass of the proton

= 1.6 \times 1 0^{- 27} kg

Answer: 2

Solution

While moving undeflected through the crossed fields, the electric and magnetic forces balance:

e E = e v B \Rightarrow E = v B

With the electric field switched off, the magnetic force alone provides the centripetal force for circular motion:

e v B = \frac{m v ^{2}}{R} \Rightarrow B = \frac{m v}{e R}

Substituting into

E = v B

E = \frac{m v ^{2}}{e R}

With

m = 1.6 \times 1 0^{- 27}

kg,

v = 2 \times 1 0^{5}

m/s,

R = 2 \times 1 0^{- 2}

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

E = \frac{( 1.6 \times 1 0 ^{- 27} ) ( 4 \times 1 0 ^{10} )}{( 1.6 \times 1 0 ^{- 19} ) ( 2 \times 1 0 ^{- 2} )} = \frac{6.4 \times 1 0 ^{- 17}}{3.2 \times 1 0 ^{- 21}} = 2 \times 1 0^{4} N/C

So x = 2.

JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Integer answer