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Two similar coils are kept mutually perpendicular such that their centres co inside. At the centre, find the ratio of the mag. field due to one coil and the resultant magnetic field by both coils, if the same current is flown.

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Explanation

$ B = B_1 = B_2 = { \mu_0 I \over 2 a } $ $B_{net} = \sqrt { B_1^2 + B_2 ^2 } $ $ = \sqrt {2B} $ $ \therefore { B \over B_{net} } = { 1 \over \sqrt 2 } $

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A long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is B. It is then bent into a circular Loop of n turns. The magnetic field at thecentre of the coil for same current will be.

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Explanation

For 1 turn $B = { \mu_0 I \over 2 r } $ Where $ l = 2 \pi r \Rightarrow r = { l \over 2 \pi } $ For n turn $B_n = { \mu_0 I \over 2r'} $ Where $ l = n (2 \pi r') $ $ B_n = \left( { \mu_0 I \over 2 { r\over n } } \right) r' = { r \over n } $ $ B_n = n^2 B $

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The mag. field due to a current carrying circular Loop of radius 3 cm at a point on the axis at a distance of 4 cm from the centre is $ 54 \mu T $ what will be its value at the centre of the LOOP.

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Explanation

$ { B Centre \over B axis } = \left( 1 + { x^2 \over a^2 } \right) ^{3/2} $ $ Bcentre = 250 \mu T $

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When the current flowing in a circular coil is doubled and the number of turns of the coil in it is halved, the magnetic field at its centre will become

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Explanation

$ B = n \left( \mu_0 I \over 2a \right) $ $ B \alpha nI $

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Two wires of same length are shaped into a square and a circle. If they carry same current, ratio of the magnetic moment is

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Explanation

Suppose length of the wire is l $ A_{square } = \left( { l \over 4 } \right) \left( {l \over 4 } \right) = { l^2 \over 16}$ $ \therefore magnetic momoent M_{square } = IA_{square } $ $ = { I l^2 \over 16 } $ $ l = 2 \pi r \Rightarrow r = { l \over 2 \pi } $ $ \therefore A_{circle} = \pi r^2 = { \pi l^2 \over 4 \pi^2 } = { l^2 \over 4 \pi } $ $ \therefore magnetic moment M_{circle} = 1 A_{circle } $ $ = { I l^2 \over 4 \pi } $ $ eq - (1) \div (2) $ $ { M_{square} \over M_{Circle }} = { \pi \over 4 } $

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Two concentric coils each of radius equal to $ 2 \pi $ cmare placed at right angles to each other. 3 Amp and 4 Amp are the currents flowing in each coil respectively. The magnetic field intensity at the centre of the coils will be …………..Tesla.

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Explanation

$ B = \sqrt { B_1^2 + B_2^2 } $ $ = {\mu_0 \over 2 r } \sqrt { I_1^2 +I_2^2 } $ $ = 5 \times 10^{-5} tesla $ . suppose point "p" is at same r distance fromthe wires

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A long solenoid has 800 turns per meter length of solenoid. A current of 1.6 Amp flows through it. The magnetic induction at the end of the solenoid on its axis is…….. tesla.

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Explanation

$ B = { \mu_o nI \over 2 } \left[ { sin 0 + sin { \pi \over 2 } } \right] = { \mu_o nI \over 2 } = 8 \times 10^{-4} tesla $

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A solenoid of 1.5 meter length and 4 cm diameter possesses 10 turn per cm. A current of 5 Amp is flowing through it. The magnetci induction at axis inside the solenoid is

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Explanation

$ n = 10 turns/ m = 1000 turns /m \Rightarrow B = \mu_o nI = 2 \pi \times 10^{-3} Tesla $

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A straight wire of length 30 cm and mass 60 milligrawm lies in a direction $ 30 ^\circ $ east of north. The earth's magnetic field at this site is horizontal and has a magnitude of 0.8 G. What current must be passed through the wire so that it may float in air ? $ ( g = 10 m/s^2 ) $

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Explanation

$ F_{mag } = F_{gravitational} $ $ BI l sin \theta = mg $ $ I = { mg \over B l sin \theta } $ $ I = 50 Amp $

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Along horizontal wire "A" carries a current of 50Amp. It is rigidly fixed. Another smallwire "B" is placed just above and parallel to "A". The weight ofwire-B per unit length is $75 \times 10^{-3} $ Newton/meter and carriesa current of 25Amp. Find the position of wire B fromAso that wire B remains suspended due to magnetic repulsion. Also indicate the direction of current in B w.r.t. to A.

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Explanation

$ {F _{mag } \over L } = { \mu_o \over 2 \pi } { I_1 I_2 \over d } $ $ {F_{mag} \over L } = {Mg \over L } $ $ { \mu_o \over 2 \pi } { I_1 I_2 \over d } = { Mg \over L } $ $ { \mu_o \over 2 \pi } { I_1 I_2 \over d } = 75 \times 10^{-3} $ $ \therefore d = {1 \over 3 } \times 10^{-2} meter $

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