QJEEM_2025_7: JEE Main 2025 Physics - Class 12 JEEM_2025_7 Solution: A proton is moving undeflected in a region of crossed electric…
A proton is moving undeflected in a region of crossed electric and magnetic fields at a constant speed of 2×10 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×10 4 N/C. The value of x is _________. Take the mass of the proton =1.6×10 −27 kg.
Step-by-Step Verified Solution
Let's break down the problem step by step:
When the proton moves undeflected in crossed electric and magnetic fields, the electric and magnetic forces balance each other. That is,
q E = q v B ,
which simplifies to
E = v B .
After the electric field is switched off, the proton moves in a circular path under the action of the magnetic force. The magnetic force provides the required centripetal force:
q v B = m v 2 r .
Solving for the magnetic field B , we get:
B = m v q r .
Now substitute this expression for B back into the equilibrium condition:
E = v B = v ( m v q r ) = m v 2 q r .
Plug in the given values:
Proton mass, m = 1.6 × 10 − 27 kg
Speed, v = 2 × 10 5 m/s
Radius, r = 2 cm = 0.02 m
Proton charge, q = 1.6 × 10 − 19 C
Thus,
E = ( 1.6 × 10 − 27 kg ) ( 2 × 10 5 m/s ) 2 ( 1.6 × 10 − 19 C ) ( 0.02 m ) .
Calculate the numerator:
( 2 × 10 5 ) 2 = 4 × 10 10 ,
So, ( 1.6 × 10 − 27 ) × ( 4 × 10 10 ) = 6.4 × 10 − 17 .
Calculate the denominator:
( 1.6 × 10 − 19 ) × ( 0.02 ) = 3.2 × 10 − 21 .
Now compute the electric field:
E = 6.4 × 10 − 17 3.2 × 10 − 21 = 2 × 10 4 N/C .
The problem states that the magnitude of the electric field is x × 10 4 N/C . Since we found
E = 2 × 10 4 N/C ,
it follows that
x = 2.