Moving Charges and Magnetism
- An electron moving in a circular orbit of radius r makes n rotations per second. The magnetic field produced at the centre has magnitude:
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Radius of circular orbit = r
No. of rotations per second = n
i.e., T = 1/n
Magnetic field at its centre, Bc =?
As we know, current i = e = e = en T (l/n)
= equivalent current Magnetic field at the centre of circular orbit, Correct Option: C
Radius of circular orbit = r
No. of rotations per second = n
i.e., T = 1/n
Magnetic field at its centre, Bc =?
As we know, current i = e = e = en T (l/n)
= equivalent current Magnetic field at the centre of circular orbit,
- A long straight wire of radius a carries a steady current I. The current is uniformly distributed over its cross-section. The ratio of the magnetic fields B and B', at radial distances a/2 and 2a respectively, from the axis of the wire is :
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For points inside the wire i.e., (r ≤ R)
Magnetic field B = μ0Ir 2πR²
For points outside the wire (r ≥ R)Magnetic field, B’ = μ0I 2πR
∴ = μ0I(a/2) = 1 : 1 B 2πa² B' μ0I 2π(2a) Correct Option: C
For points inside the wire i.e., (r ≤ R)
Magnetic field B = μ0Ir 2πR²
For points outside the wire (r ≥ R)Magnetic field, B’ = μ0I 2πR
∴ = μ0I(a/2) = 1 : 1 B 2πa² B' μ0I 2π(2a)
- A current carrying coil is subjected to a uniform magnetic field. The coil will orient so that its plane becomes
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The plane of coil will orient itself so that area vector aligns itself along the magnetic field.So, the plane will orient perpendicular to the magnetic field.
Correct Option: D
The plane of coil will orient itself so that area vector aligns itself along the magnetic field.So, the plane will orient perpendicular to the magnetic field.
- A uniform magnetic field acts at right angles to the direction of motion of electron. As a result, the electron moves in a circular path of radius 2cm. If the speed of electron is doubled, then the radius of the circular path will be
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r = mv or r ∝ v qB
As v is doubled, the radius also becomes double. Hence, radius = 2 × 2 = 4 cmCorrect Option: C
r = mv or r ∝ v qB
As v is doubled, the radius also becomes double. Hence, radius = 2 × 2 = 4 cm
- 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
