Question

a charged particle with velocity 2 ×10^ 3 metre per second passes undeflected through the electric field and magnetic fields in mutually perpendicular direction.The magnetic field is 1.5 Tesla, the magnitude of electric field will be

1) 1.5×10^3 N/C

2)2×10^3 N/C

3)3×10^3 N/C

4)1.33×10^3 N/C​

Answer 1

Answer:

Given Information:

• Velocity = 2 × 10³ m/s
• Magnetic Field = 1.5 T
• Electric Field = ?

According to the question, the charged particle goes undeflected. Hence the Electric Force and Magnetic Force acting on the particle is equal in magnitude.

Electric Force = Charge × Electric Field

→ F = q.E   ... ( 1 )

Magnetic Force = Charge × (Cross product of Velocity and Magnetic Field)

→ F = q.(v × B)   ... ( 2 )

Now equating both ( 1 ) and ( 2 ) we get:

→ q.E = q.(v × B)

→ E = v × B

According to the question, v is 2 × 10³ m/s and B is 1.5 T. Therefore Electric Field is:

→ E = 2 × 10³ × 1.5  

→ E = 3 × 10³ N/C

Hence Option (3) is the correct option.