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Which of the following gives the value of the magnetic field according to, Biot-Savart’s law :

$\frac{i ∆ lsin (θ)}{r^{2}}$

$\frac{μ_{0}}{4 π} \frac{i ∆ lsin (θ)}{r}$

$\frac{μ_{0}}{4 π} \frac{i ∆ lsin (θ)}{r^{2}}$

$\frac{μ_{0}}{4 π} i ∆ lsin (θ)$

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Explanation

$M a g n e t i c f i e l d a c c o r d i n g t o, B i o t - S a v a r t ’ s l a w = \frac{μ_{0}}{4 π} \frac{i ∆ l \sin (θ)}{r^{2}}$

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The earth’s magnetic field at a given point is $0.5 \times 10^{- 5} W b - m^{- 2} .$ This field is to be annulled by magnetic induction at the center of a circular conducting loop of radius 5.0cm. The current required to be flown in the loop is nearly :

0.2 A

0.4A

40A

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Explanation

(b) Magnetic field at the centre of circular loop

$B = \frac{μ_{0}}{4 π} \frac{2 πi}{r} \Rightarrow 0.5 \times 10^{- 5} = \frac{10^{- 7} \times 2 \times 3.14 \times i}{5 \times 10^{- 2}}$

$i = 0.4 A$

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The magnetic moment of a current (i) carrying circular coil of radius (r) and number of turns (n) varies as :

$1 / r^{2}$

$1 / r$

$r^{2}$

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Explanation

(d) $M = n i A = n i ({πr}^{2}) \Rightarrow M \propto r^{2}$

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The current is flowing in the south direction along a power line. The direction of the magnetic field above the power line (neglecting earth's field) is :

South

East

North

West

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Explanation

According to the right-hand thumb rule, the direction of the magnetic field above a straight conductor carrying current from south to north will be towards the west. This rule provides a convenient way to determine the direction of the magnetic field around a current-carrying wire.

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For the magnetic field to be maximum due to a small element of current carrying conductor at a point, the angle between the element and the line joining the element to the given point must be :

0°

90°

180°

45°

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Explanation

(b) The magnetic field due to small element of conductor of length is given by $d B = \frac{μ_{0}}{4 π} \frac{I d I \sin θ}{r^{2}}$
This value will be maximum when
$\sin θ = 1 = \sin 90 °$ or, $θ = 90 °$

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An electron and a proton enter a magnetic field perpendicularly. Both have the same kinetic energy. Which of the following is true:

Trajectory of electron is less curved

Trajectory of proton is less curved

Both trajectories are equally curved

Both move on a straight-line path

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Explanation

(b) We have $q v B = \frac{m v^{2}}{r}$ or $r = \frac{m v}{q B} = \frac{\sqrt{2 m K}}{q B}$
For same kinetic energy K, we have $r \propto \sqrt{m}$
Hence path of proton will have larger r and is therefore less curved.

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An electron and a proton with equal momentum enter perpendicularly into a uniform magnetic field, then :

The path of proton shall be more curved than that of electron

The path of proton shall be less curved than that of electron

Both are equally curved

Path of both will be a straight line

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Explanation

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One proton beam enters a magnetic field of $10^{- 4} T$ normally, Specific charge = $10^{11} C / k g .$ velocity = $10^{7} m / s .$ What is the radius of the circle described by it:

$0.1 m$

$1 m$

$10 m$

None of these

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Explanation

(b) $r = \frac{m v}{q B} = \frac{10^{7}}{10^{11} \times 10^{- 4}} = 1 m (∵ q / m = 10^{11} C / k g)$

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In a cyclotron, the angular frequency of a charged particle is independent of :

Mass

Speed

Charge

Magnetic field

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Explanation

$ω = \frac{2 π}{T} = \frac{q B}{m} (∵ T = \frac{2 πm}{q B})$

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Maximum kinetic energy of the positive ion in the cyclotron is :

$\frac{q^{2} B r_{0}}{2 m}$

$\frac{q B^{2} r_{0}}{2 m}$

$\frac{q^{2} B^{2} r_{0}^{2}}{2 m}$

$\frac{q B r_{0}}{2 m^{2}}$

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Explanation

$\Rightarrow K_{m a x} = \frac{1}{2} m {(\frac{q B r_{0}}{m})}^{2} = \frac{q^{2} B^{2} r_{0}^{2}}{2 m}$

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