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A particle mass m, charge Q, and kinetic energy T enter a transverse uniform magnetic field of induction $\vec{B}$ . After 3sec the kinetic energy of the particle will be :

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Explanation

After passing through a magnetic field, the magnitude of its mass and velocity of the particle remain the same, so its energy does not change, i.e., kinetic energy will remain T.

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A galvanometer of resistance 50 $Ω$ is connected to a battery of 3 V along with a resistance of 2950 $Ω$ in series. A full-scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be

5050 $Ω$

5550 $Ω$

6050 $Ω$

4450 $Ω$

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Explanation

To reduce the deflection in a galvanometer, the total resistance in the circuit needs to be increased. By calculating the required total resistance using the formula (V/I) = R, where V is the battery voltage, I is the new desired current, and R is the total resistance, the resistance to be added in series can be determined.

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In Bainbridge mass spectrograph a potential difference of 1000 V is applied between two plates distant 1 cm apart and magnetic field in B = 1T. The velocity of undeflected positive ions in m/s from the velocity selector is

$10^{7} m / s$

$10^{4} m / s$

$10^{5} m / s$

$10^{2} m / s$

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Explanation

(c)

$v = \frac{E}{B}; where E = \frac{V}{d} = \frac{1000}{1 \times 10^{- 2}} = 10^{5} V / m \Rightarrow v = \frac{10^{5}}{1} = 10^{5} m / s$

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An $α$ particle is accelerated through a p.d of $10^{6}$ volt then K.E. of particle will be

8 MeV

4 MeV

2 MeV

1 MeV

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Explanation

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A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the centre of the loop is B. It is then bent into a
circular coil of n turns. The magnetic field at the centre of this coil of n turns will be:

n²B

2nB

2n²B

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Explanation

Length of wire = perimeter of the circle l = 2π R (R is radius of loop of one turn)
now it is again bend in n turns of radius r
so 2π R = nx2π r
r = R/n
For 1 turn coil B at the centre is given by
B = µ₀i/R
For n turn coil B at the centre is given by
B’ = n µ₀i/r
= n µ₀i/R/n = n²( µ₀i/R) = n² B

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A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in the equilibrium state. The energy required to rotate it by 60^o is W. Now the torque required to keep the magnet in this new position is:

$\frac{W}{\sqrt{3}}$

$\sqrt{3} W$

$\frac{\sqrt{3} W}{2}$

$\frac{2 W}{\sqrt{3}}$

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Explanation

Torque = MBsinθ
= MB sin $60^{0}$ -------(1)
Work done in displacing the magnet from an angle $θ_{1}$ to $θ_{2}$ is -
W = MB(cos $θ_{1}$ – cos $θ_{2}$ )
W = MB(1 – cos $60^{0}$ ) -------(2)
From (1) and (2)
Torque= $\frac{\frac{\sqrt{3} W}{2}}{\frac{1}{} 2} = \sqrt{3} W$

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An electron is moving in a circular path under the influence of a transverse magnetic field of 3.57 x 10^-2 T. If the value of e/m is 1.76 x 10¹¹ C/kg, the frequency of revolution of the electron is:

1 GHz

100 MHz

62.8 MHz

6.28 MHz

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Explanation

When an electron enters magnetic field it will acquire circular path. Its radius is given by
r = mv/qB
Time table for complete cycle
T = 2 πr/v
frequency = 2 π/T = eB/m
F = eB/2 πm
= 1.76 x10¹¹x 3.57x 10 ^-2 /2 π
= 1 GHz

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A 250-Turn rectangular coil of length 2.1 cm and width 1.25 cm carries a current of 85 $μ$ and subjected to the magnetic field of strength 0.85 T. Work done for rotating the coil by 180° against the torque is:

$4.55 μ J$

$2.3 μ J$

$1.15 μ J$

$9.8 μ J$

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Explanation

Given a rectangular coil of length 2.1 cm and width 1.25 cm
Current through coil = 85 $μ A$
No. of turns = 250
B= 0.85 T
Work done, W= MB $(\cos θ_{1} - \cos θ_{2})$
When it is rotated by angle 180⁰ then
W= MB $(\cos 0^{0} - \cos 180^{0}) = M B (1 + 1) = 2 M B$
W = 2(NIA)B
W= $2 \times 250 \times 85 \times 10^{- 6} [1.25 \times 2.1 \times 10^{- 4}] \times 85 \times 10^{- 2}$

W= $9.8 μ J$

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A metallic rod of mass per unit length 0.5 kg m^-1 is lying horizontally on a smooth inclined plane which makes an angle of 30° with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction 0.25 T is acting on it in the vertical direction. The current flowing in the rod to keep stationery is:

7.14 A

5.98 A

14.76 A

11.32 A

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Explanation

The force required to keep the rod stationary on an inclined plane is equal to the weight component along the plane. This force is provided by the magnetic force on the current-carrying rod. Using the formula F = BILsinÎ¸, where B is the magnetic field, I is the current, L is the length of the rod, and Î¸ is the angle of inclination, the current required can be calculated.

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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 30^o with the direction of the field, the torque required to keep the coil in stable equilibrium will be:

0.15 Nm

0.20 Nm

0.24 Nm

0.12 Nm

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