Physics Moving Charges and Magnetism
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A wire , bent in the shape of an arc of a circle, carrying a current of and having radius and another wire , also bent in the shape of arc of a circle, carrying a current of and having radius of , are placed as shown in the figure. The ratio of the magnetic fields due to the wires and B at the common centre 0 is:
A wire , bent in the shape of an arc of a circle, carrying a current of and having radius and another wire , also bent in the shape of arc of a circle, carrying a current of and having radius of , are placed as shown in the figure. The ratio of the magnetic fields due to the wires and B at the common centre 0 is:
If momentum (P), area (A) and time (T) are taken to be the fundamental quantities, then the dimensional formula for energy is :
If momentum (P), area (A) and time (T) are taken to be the fundamental quantities, then the dimensional formula for energy is :
E = P? A? T? [ML²T? ²] = [MLT? ¹]? [L²]? [T]?
a=1, 2b+a=2⇒b=1/2, -a+c=-2⇒c=-1. [PA¹/²T? ¹].
The figure shows a region of length 'l' with a uniform magnetic field of 0.3 T in it and a proton entering the region with velocity 4 × 10⁵ ms⁻¹ making an angle 60° with the field. If the proton completes 10 revolution by the time it cross the region shown, 'l' is close to (mass of proton = 1.67 × 10⁻²⁷ kg, charge of the proton = 1.6 × 10⁻¹⁹C)
The figure shows a region of length 'l' with a uniform magnetic field of 0.3 T in it and a proton entering the region with velocity 4 × 10⁵ ms⁻¹ making an angle 60° with the field. If the proton completes 10 revolution by the time it cross the region shown, 'l' is close to (mass of proton = 1.67 × 10⁻²⁷ kg, charge of the proton = 1.6 × 10⁻¹⁹C)
l = 10 × pitch = 10 × (2πmvcosθ/qB)
= 20π × 1.67×10? ²? × 4×10? × (1/2) / (1.6×10? ¹? × 0.3) = 0.44 m
A beam of protons with speed 4 × 10⁵ ms⁻¹ enters a uniform magnetic field of 0.3 T at an angle of 60° to the magnetic field. The pitch of the resulting helical path of protons is close to: (Mass of the proton = 1.67 × 10⁻²⁷ kg, charge of the proton = 1.69 × 10⁻¹⁹ C)
A beam of protons with speed 4 × 10⁵ ms⁻¹ enters a uniform magnetic field of 0.3 T at an angle of 60° to the magnetic field. The pitch of the resulting helical path of protons is close to: (Mass of the proton = 1.67 × 10⁻²⁷ kg, charge of the proton = 1.69 × 10⁻¹⁹ C)
Pitch = (Vcosθ)T = (Vcosθ)2πm/eB
= (4×10? cos60°) (2π (1.67×10? ²? )/ (0.3×10? ¹? ×1.69×10¹? )
= 4 cm
An electron is moving along +x direction with a velocity of 6 × 10? ms?¹. It enters a region of uniform electric field of 300 V/cm pointing along +y direction. The magnitude and direction of the magnetic field set up in this region such that the electron keeps moving along the x direction will be:
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An electron is moving along +x direction with a velocity of 6 × 10? ms?¹. It enters a region of uniform electric field of 300 V/cm pointing along +y direction. The magnitude and direction of the magnetic field set up in this region such that the electron keeps moving along the x direction will be:
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A particle of charge q and mass m is moving with a velocity -vi (v ≠ 0) towards a large screen placed in the Y-Z plane at a distance d. If there is a magnetic field B = B?k, the minimum value of v for which the particle will not hit the screen is:
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A particle of charge q and mass m is moving with a velocity -vi (v ≠ 0) towards a large screen placed in the Y-Z plane at a distance d. If there is a magnetic field B = B?k, the minimum value of v for which the particle will not hit the screen is:
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r = mv/qB?
To not collide, r
Note: It should be maximum instead of minimum.
