Moving Charges and Magnetism
- A charge moving with velocity v in X-direction is subjected to a field of magnetic induction in negative X-direction. As a result, the charge will
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The force acting on a charged particle in magnetic field is given by
→F = q (→v × →B) or F = qvB sin θ,
When angle between v and B is 180°, F = 0Correct Option: A
The force acting on a charged particle in magnetic field is given by
→F = q (→v × →B) or F = qvB sin θ,
When angle between v and B is 180°, F = 0
- A beam of electrons is moving with constant velocity in a region having simultaneous perpendicular electric and magnetic fields of strength 20 Vm-1 and 0.5 T respectively at right angles to the direction of motion of the electrons. Then the velocity of electrons must be
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The electron moves with constant velocity without deflection. Hence, force due to magnetic field is equal and opposite to force due to electric field.
qvB = qE ⇒ v = E = 20 = 40 m/s B 0.5 Correct Option: C
The electron moves with constant velocity without deflection. Hence, force due to magnetic field is equal and opposite to force due to electric field.
qvB = qE ⇒ v = E = 20 = 40 m/s B 0.5
- An electron enters a region where magnetic field (B) and electric field (E) are mutually perpendicular, then
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When the deflection produced by electric field is equal to the deflection produced by magnetic field, then the electron can go undeflected.
Correct Option: D
When the deflection produced by electric field is equal to the deflection produced by magnetic field, then the electron can go undeflected.
- Two identical long conducting wires AOB and COD are placed at right angle to each other, with one above other such that ‘O’ is their common point for the two. The wires carry I1 and I2 currents respectively. Point ‘P’ is lying at distance ‘d’ from ‘O’ along a direction perpendicular to the plane containing the wires. The magnetic field at the point ‘P’ will be :
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Net magnetic field, B = √B1² + √B2²
= 
μ0I1 
² + μ0I2 ² 2πd 2πd ∵ B1 = μ0I1 And B2 = μ0I2 2πd 2πd = μ0 √I1² + I2² 2πd Correct Option: D
Net magnetic field, B = √B1² + √B2²
= μ0I1 ² + μ0I2 ² 2πd 2πd ∵ B1 = μ0I1 And B2 = μ0I2 2πd 2πd = μ0 √I1² + I2² 2πd
- A deuteron of kinetic energy 50 keV is describing a circular orbit of radius 0.5 metre in a plane perpendicular to the magnetic field B. The kinetic energy of the proton that describes a circular orbit of radius 0.5 metre in the same plane with the same B is
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For a charged particle orbiting in a circular path in a magnetic field
mv² = Bqv ⇒ v = Bqr r m
or, mv² = Bqvr
Also, EK = 1 mv² = 1 Bqvr = Bq. r . Bqr = B²q²r² 2 2 2 m 2m For deuteron, E1 = B²q²r² 2 × 2m For proton, E2 = B²q²r² 2m = E1 = 1 ⇒ 50 KeV = 1 ⇒ E2 = 100 keV E2 2 E2 2 Correct Option: D
For a charged particle orbiting in a circular path in a magnetic field
mv² = Bqv ⇒ v = Bqr r m
or, mv² = Bqvr
Also, EK = 1 mv² = 1 Bqvr = Bq. r . Bqr = B²q²r² 2 2 2 m 2m For deuteron, E1 = B²q²r² 2 × 2m For proton, E2 = B²q²r² 2m = E1 = 1 ⇒ 50 KeV = 1 ⇒ E2 = 100 keV E2 2 E2 2
