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A galvanometer of resistance, G is shunted by a resistance S ohm. To keep the main current in the circuit unchanged, the resistance to be put in series with the galvanometer is

$\frac{S^{2}}{(S + G)}$

$\frac{S G}{(S + G)}$

$\frac{G^{2}}{(S + G)}$

$\frac{G}{(S + G)}$

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Charge q is uniformly spread on a thin ring of radius R. The ring rotates about its axis with a uniform frequency f Hz. The magnitude of magnetic induction at the center of the ring is :

$\frac{μ_{0} q f}{2 R}$

$\frac{μ_{0} q}{2 f R}$

$\frac{μ_{0} q}{2 π f R}$

$\frac{μ_{0} q f}{2 πR}$

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Explanation

We know

$B = \frac{μ_{0} i}{2 R}$

$q = it \Rightarrow i = \frac{q}{t} = q f [∵ f = \frac{1}{t}]$

$B = \frac{μ_{0} q f}{2 R}$

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A thin ring of radius R metre has charge q coulomb uniformly spread on it.The ring rotates about its axis with a constant frequency of f revolution/s.The value of magnetic induction in $W b m^{- 2}$ at the centre of the ring is

$\frac{μ_{0} q f}{2 πR}$

$\frac{μ_{0} q}{2 π f R}$

$\frac{μ_{0} q}{2 f R}$

$\frac{μ_{0} q f}{2 R}$

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Explanation

The magnetic field at the centre of the circle

$= \frac{μ_{0}}{4 π} \times \frac{2 πi}{R} = \frac{μ_{0}}{2} \frac{(q)}{R t}$

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A galvanometer has a coil of resistance $100 Ω$ and gives a full scale deflection for 30 mA current.If it is to work as a voltmeter of 30V range, the resistance required to be added will be

900 $Ω$

1800 $Ω$

500 $Ω$

1000 $Ω$

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Explanation

Required resistance R= $\frac{V}{i_{g}} - G$

$= \frac{30}{30 \times 10^{- 3}} - 100 = 900 Ω$

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A square current carrying loop is suspended in a uniform magnetic field acting in the plane of the loop. If the force on one arm of the loop is $\vec{F}$ , the net force on the remaining three arms of the loop is

3 $\vec{F}$

$-$ $\vec{F}$

$-$ 3 $\vec{F}$

$\vec{F}$

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Explanation

When a current-carrying loop is placed in a uniform magnetic field, the force experienced by each arm of the loop due to the magnetic field is equal in magnitude but opposite in direction. Therefore, the net force on the remaining three arms of the loop is equal in magnitude but opposite in direction to the force on the first arm.

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A closely wound solenoid of 2000 turns and area of cross-section $1.5 \times 10^{- 4} m^{2}$ carries a current of $2.0 A .$ It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field $5 \times 10^{- 2} T$ making an angle of $30 °$ with the axis of the solenoid. The torque on the solenoid will be

$3 \times 10^{- 3} N ‐ m$

$1.5 \times 10^{- 3} N ‐ m$

$1.5 \times 10^{- 2} N ‐ m$

$3 \times 10^{- 2} N ‐ m$

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Explanation

$Given, N = 2000, A = 1.5 \times 10^{- 4} m^{2} i = 2.0 A B = 5 \times 10^{- 2} T, and θ = 30 ° Torque, τ = N i B A \sin θ = 2000 \times 2 \times 5 \times 10^{- 2} \times 1.5 \times 10^{- 4} \times \sin 30 ° = 2000 \times 50 \times 10^{- 6} \times \frac{1}{2} = 1.5 \times 10^{- 2} Nm$

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A particle having a mass of $10^{- 2} kg$ carries a charge of $5 \times 10^{- 8} C .$ The particle is given an initial horizontal velocity of $10^{5} {ms}^{- 1}$ in the presence of electric field $\vec{E}$ and magnetic field $\vec{B}$ . To keep the particle moving in a horizontal direction, it is necessary that

$\vec{B}$ should be perpendicular to the direction of velocity and $\vec{E}$ should be along the direction of velocity.

Both $\vec{B}$ and $\vec{E}$ should be along the direction of velocity.

Both $\vec{B}$ and $\vec{E}$ are mutually perpendicular and perpendicular to the direction of velocity.

$\vec{B}$ should be along the direction of velocity and $\vec{E}$ should be perpendicular to the direction of velocity.

Which one of the following pairs of statements are possible?

1. (1) and (3) 2. (3) and (4)

3. (2) and (3) 4. (2) and (4)

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Explanation

Both $\vec{B}$ and $\vec{E}$ should be along the direction of velocity and both $\vec{B}$ and $\vec{E}$ are mutually perpendicular and perpendicular to the direction of velocity.

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Under the influence of a uniform magnetic field, a charged particle moves with constant speed v in a circle of radius R. The time period of rotation of the particle -

depends on v and not on R

depends on R and not on v

is independent of both v and R

depends on both v and R

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Explanation

The time period of circular motion of the charged particle is given by

$T = \frac{2 πr}{v} = \frac{2 π}{v} \times \frac{mv}{Bq} or T = \frac{2 πm}{Bq}$

Hence, time period of rotation of the charged particle in uniform magnetic field is independent of both v and R.

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The magnetic force acting on a charged particle of charge $- 2 μ C$ in a magnetic field of 2T acting in y-direction, when the particle velocity is $(2 \hat{i} + 3 \hat{j}) \times 10^{6} m s^{- 1}$ is.

8 N in -z-direction

4 N in z-direction

8 N in y-direction

8 N in z-direction

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Explanation

When a charge q moves with velocity $\vec{v}$ inside a magnetic field of strength $\vec{B,}$ then force on the charge is called magnetic Lorentz force. The magnetic Lorentz force is in the direction of vector $\vec{v} \times \vec{B}$ .

Magnetic Lorentz force $\vec{F} = q (\vec{v} \times \vec{B})$

= $- 2 \times 10^{- 6} [2 \times 2 \times 10^{6}]$

= 8 N along negative z-axis

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A galvanometer having a coil resistance of $60 Ω$ shows full scale deflection when a current of 1.0A passes through it. It can be converted into an ammeter to read currents upto 5.0A by

putting in series resistance of 240 $Ω$

putting in parallel resistance of 240 $Ω$

putting in series resistance of 15 $Ω$

putting in parallel resistance of 15 $Ω$

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Explanation

To convert a galvonometer to ammeter a small resistance is connected in parallel to the coil of the galvonometer.

Here, $G_{1} = 60 Ω, I_{g} = 1.0 A, I = 5 A$

$I_{g} G_{1} = (I - I_{g}) S$

$S = \frac{I_{g} G_{1}}{I - I_{g}} = \frac{1}{5 - 1} \times 60 = 15 Ω$

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