Moving perpendicular to field B, a proton and an alpha particle both enter an area of uniform magnetic field B. If the kinetic energy of the proton is 1 MeV and the radius of the circular orbits for both particles is equal, the energy of the alpha particle will be:

1. 4 MeV

2. 0.5 MeV

3. 1.5 MeV

4. 1 MeV

Subtopic:  Lorentz Force |
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A rectangular coil of length 0.12 m and width 0.1 m having 50 turns of wire is suspended vertically in a uniform magnetic field of strength 0.2 Wb/m2. The coil carries a current of 2 A. If the plane of the coil is inclined at an angle of 30o with the direction of the field, the torque required to keep the coil in stable equilibrium will be:

1. 0.15 N-m

2. 0.20 N-m

3. 0.24 N-m

4. 0.12 N-m

Subtopic:  Current Carrying Loop: Force & Torque |
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A wire carrying current I has the shape as shown in the adjoining figure. Linear parts of the wire are very long and parallel to X-axis while the semicircular portion of radius R is lying in the Y –Z plane. The magnetic field at point O is: 1. $$B=\frac{\mu i }{4\pi R}\left ( \pi \hat{i}+2\hat{k} \right )$$
2. $$B=-\frac{\mu i }{4\pi R}\left ( \pi \hat{i}-2\hat{k} \right )$$
3. $$B=-\frac{\mu i }{4\pi R}\left ( \pi \hat{i}+2\hat{k} \right )$$
4. $$B=\frac{\mu i }{4\pi R}\left ( \pi \hat{i}-2\hat{k} \right )$$

Subtopic:  Magnetic Field due to various cases |
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An electron moving in a circular orbit of radius r makes n rotations per second. The magnetic field produced at the centre has a magnitude:

1. μ0ne/2πr

2. zero

3. n2e/r

4. μ0ne/2r

Subtopic:  Magnetic Field due to various cases |
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In an ammeter, 0.2% of the main current passes through the galvanometer. If the resistance of the galvanometer is G, the resistance of the ammeter will be:

1. $\frac{1}{499}G$

2. $\frac{499}{500}G$

3. $\frac{1}{500}G$

4. $\frac{500}{499}G$

Subtopic:  Conversion to Ammeter & Voltmeter |
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Two identical long conducting wires AOB and COD are placed at a right angle to each other, with one above the other such that 'O' is the common point for the two. The wires carry I1 and I2 currents, respectively. Point 'P' is lying at a distance 'd' from 'O' along a direction perpendicular to the plane containing the wires. What will be the magnetic field at the point "P"?

1. $\frac{{\mu }_{0}}{2\mathrm{\pi d}}\left(\frac{{I}_{1}}{{I}_{2}}\right)$

2. $\frac{{\mu }_{0}}{2\mathrm{\pi d}}\left({I}_{1}+{I}_{2}\right)$

3. $\frac{{\mu }_{0}}{2\mathrm{\pi d}}\left({I}_{1}^{2}+{I}_{2}^{2}\right)$

4. $\frac{{\mu }_{0}}{2\mathrm{\pi d}}{\left({I}_{1}^{2}+{I}_{2}^{2}\right)}^{1/2}$

Subtopic:  Magnetic Field due to various cases |
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A current loop in a magnetic field:
 1 can be in equilibrium in one orientation. 2 can be in equilibrium in two orientations, both the equilibrium states are unstable. 3 can be in equilibrium in two orientations, one stable while the other is unstable. 4 experiences a torque whether the field is uniform or non-uniform in all orientations.
Subtopic:  Magnetic Field due to various cases | Current Carrying Loop: Force & Torque |
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When a proton is released from rest in a room, it starts with an initial acceleration a0 towards the east. When it is projected towards the north with a speed of v0, it moves with an initial acceleration of 3a0 towards the east. What are the electric and magnetic fields in the room?

1. $$\frac{Ma_{0}}{e}~west,~\frac{Ma_{0}}{ev_{0}}~up$$
2. $$\frac{Ma_{0}}{e}~west,~\frac{2Ma_{0}}{ev_{0}}~down$$
3. $$\frac{Ma_{0}}{e}~east,~\frac{2Ma_{0}}{ev_{0}}~up$$
4. $$\frac{Ma_{0}}{e}~east,~\frac{3Ma_{0}}{ev_{0}}~down$$

Subtopic:  Lorentz Force |
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Two similar coils of radius R are lying concentrically with their planes at right angles to each other. The currents flowing in them are I and 2I, respectively. What will be the resultant magnetic field induction at the centre?

1. $\frac{\sqrt{5}{\mu }_{0}I}{2R}$

2. $\frac{3{\mu }_{0}I}{2R}$

3. $\frac{{\mu }_{0}I}{2R}$

4. $\frac{{\mu }_{0}I}{R}$

Subtopic:  Magnetic Field due to various cases |
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A millivoltmeter of 25 mV range is to be converted into an ammeter of 25 A range. The value (in ohm) of necessary shunt will be:

1. 0.001

2. 0.01

3. 1

4. 0.05

Subtopic:  Conversion to Ammeter & Voltmeter |
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