Q30. An electron enters a crossed electrical and magnetic field region with a speed of 100 m s-¹ alo

Q: What is the correct answer for CUET UG Physics Q30?

The correct answer is a. For a charged particle to pass undeviated through crossed electric (E) and magnetic (B) fields, the electric force (F_e = qE) and magnetic force (F_m = q(v × B)) must be equal in magnitude and opposite in direction. This is the principle of a velocity selector, where the net force is zero.So, |qE| = |q(v × B)|, which simplifies to E = vB.Given: speed (v) = 100 m/s, magnetic field strength (B) = 5.

Q: Q30. An electron enters a crossed electrical and magnetic field region with a speed of 100 m s-¹ along + x-axis and passes undeviated. The strength of magnetic field is 5.0 T and it acts along +…

The correct answer is a. For a charged particle to pass undeviated through crossed electric (E) and magnetic (B) fields, the electric force (F_e = qE) and magnetic force (F_m = q(v × B)) must be equal in magnitude and opposite in direction. This is the principle of a velocity selector, where the net force is zero.So, |qE| = |q(v × B)|, which simplifies to E = vB.Given: speed (v) = 100 m/s, magnetic field strength (B) = 5.

Q: What are the options for this CUET UG Physics question?

A. (500 V m⁻¹) · B. -(500 V m⁻¹)k · C. (100 V m⁻¹) · D. -(100 V m⁻¹)k

Question

Q30. An electron enters a crossed electrical and magnetic field region with a speed of 100 m s-¹ along + x-axis and passes undeviated. The strength of magnetic field is 5.0 T and it acts along + z-axis. Give the electrical field in vector notations indicating its magnitude as well as direction:
(a) (500 V m⁻¹)
(b) -(500 V m⁻¹)k
(c) (100 V m⁻¹)
(d) -(100 V m⁻¹)k

FirstInTest · Accepted Answer

For a charged particle to pass undeviated through crossed electric (E) and magnetic (B) fields, the electric force (F_e = qE) and magnetic force (F_m = q(v × B)) must be equal in magnitude and opposite in direction. This is the principle of a velocity selector, where the net force is zero.So, |qE| = |q(v × B)|, which simplifies to E = vB.Given: speed (v) = 100 m/s, magnetic field strength (B) = 5.0 T.E = (100 m/s) * (5.0 T) = 500 V/m.For the direction: The electron (charge q = -e) moves along the +x-axis (v = v î). The magnetic field is along the +z-axis (B = B k̂).Magnetic force F_m = q(v × B) = (-e)(v î × B k̂) = -e v B (î × k̂) = -e v B (-ĵ) = e v B ĵ.So, the magnetic force is in the +y direction.For the electron to pass undeviated, the electric force (F_e = qE = -eE) must be opposite to the magnetic force, meaning F_e must be in the -y direction.So, -eE = - (e v B ĵ) => E = v B ĵ.Thus, the electric field is in the +y direction, with a magnitude of 500 V/m. Option (a) provides the correct magnitude.

Answer

(500 V m⁻¹)

Answer

-(500 V m⁻¹)k

Answer

(100 V m⁻¹)

Answer

-(100 V m⁻¹)k

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