NEET Physics Questions: Magnetic Effects of Current and Magnetism

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An element $ \vec dl = dx \uparrow $ (where dx = 1 cm) is placed at the origin and carries a large current I = 10 Amp. What is the mag. field on the Y-axis at a distance of 0.5 meter ?




$ d l = dx = 10^{-2} m $ I = 10 Amp ; r =0.5 m $ d \vec B = { \mu_0 \over 4 \pi } { I \vec dl \times \vec r \over r^3 } $ $ = 4 \times 10^{-8} \hat k\; tesla $

Two straight long conductors AOB and COD are perpendicular to each other and carry currents $I_1$ and $I_2$. The magnitude of the mag. field at a point "P" at a distance "a" from the point "O" in a direction perpendicular to the plane ABCD is





Point "P" is lying symmetricallyw.r.t.the two long wires $ B_1 = { \nu_0 \over 2 \pi } { I_1 \over a } ;$ $ B_2 = { \mu_0 \over 2 \pi} { I_2 \over a } $ $ B = \sqrt { B_1^2 + B_1^2 } $ $ = { \mu_0 \over 2 \pi a } ( I_1^2 + I_2^2 )^{1/2} $

A length L of wire carries a steady current 1. It is bent first to form a coil of 1 turn. The same length is now bent more sharply to give a double loop of smaller radius. The magnetic field at the centre caused by the same current is.................




$ B = N \left( {\mu_0 I \over 2 \pi } \right) \Rightarrow B \alpha {N \over r } \Rightarrow {B_1 \over B_2 } = { N_1 \over r_1} \times { r_2 \over N_2 }$ $ B_2 = 4 B_1 $ Technique $ B_2 = n^2 B1 = (2)^2 B1 = 4B1 $
If a long hollow copper pipe carries a direct current, the magnetic field associated with the current will be ................




only outside the pipe Hollow copper pipe I = 0 $ \therefore B = { \mu_0 \over 2 \pi } { I \over r } = 0 $ i.e. Inside mag.fieldis ZERO
The magnetic induction at a point P which is at a distance 4 cm from a long current carrying wire is $10^{-8} $ tesla. The field of induction at a distance 12 cm from the same current would be ...............tesla.




$ B = { \mu_0 \over 2 \pi } { I \over y } \Rightarrow { B_1 \over B_2 } = { y_2 \over y_1 } $ $ \therefore B \alpha { 1 \over y } $ $ { 10^{-8} \over B_2 } = { 12 \times 10^{-2} \over 4 \times 10^{-2}} $ $ B_2 = 3.33 \times 10^{-9} tesla $
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The strength of the magnetic field at a point y near a long straight current carrying wire is B. The field at a distance y/2 will be




$ B = { \mu_0 \over 2 \pi } { I \over y } \Rightarrow B \alpha { 1 \over y } \Rightarrow { B_1 \over B_2} = { y_2 \over y_1 } $ $ \Rightarrow B_2 = 2B_1 $
The mag. field (B) at the centre of a circular coil of radius "a", through which a current I flows is.............




$ B = { \mu_0 \over 2 } { I \over a } \Rightarrow B \alpha I $
A current of a 1 Amp is passed through a straight wire of length 2 meter. The magnetic field at a point in air at a distance of 3 meters from either end of wire and lying on the axis of wire will be............




$ \theta_1 = \theta_2 = 0 $ $ B = { \mu_0 \over 4 \pi } { I \over y } [ sin \theta_1 + sin \theta_2 ] $
If the strength of the magnetic field produced at 10 cm away from a infinitely long straight conductor is $10^{-5} $ tesla. The value of the current flowing in the conductor will be............... Ampere.




I = 5 Amp
A long straight wire of radius "a" carries a steady current I the current is uniformly distributed across its cross-section. The ratio of the magnetic field at a/2 and 2a is .................




$ ( B_{wire } ) _{r \lt 9} = ( B_{wire } ) _{r \gt 9}$ $ \left( { \mu_0 I \over 2 \pi 9^2 } \right) { 9 \over 2 } = { \mu_0 \over 2 \pi } { I \over 29 } $ $B_1 = B_2 $ ${ B_1 \over B_2 } = 1 $

At a distance of 10 cm from a long straight wire carrying current, the magnetic field is $ 4 \times 10^{-2} $ .At the distance of 40 cm, the magnetic field will be Tesla.





$ B = { \mu_o \over 2 \pi } { I \over y } \Rightarrow B \alpha { 1 \over y } \Rightarrow { B_1 \over B_2 } = {y_2 \over y_1} \Rightarrow B_2 = 1 \times 10^{-2} $

A He nucleus makes a full rotation in a circle of radius 0.8 meter in 2 sec. The value of the mag. field B at the centre of the circle will be ………...Tesla.




$ I = { Q \over t } = { 2 \times 1.6 \times 10^{-19 } \over 2 } = 1.6 \times 10^{-19} Amp $ $ B = { \mu_0 \over 2 a } = \mu_0 \times 10^{-19} Tesla $
The direction of mag. field lines close to a straight conductor carrying current will be ...............




The magnetic field lines around a straight conductor carrying current form concentric circles in a plane perpendicular to the length of the conductor. This is described by the Right Hand Thumb Rule: if you grasp the conductor with your right hand, with the thumb pointing in the direction of the current, your fingers will curl in the direction of the magnetic field lines.
Due to 10 Amp of current flowing in a circular coil of 10 cmradius, the mag. field produced at its centre is$ \ pi \times 10^{-3 } $ Tesla. The number of turns in the coil will be




$ B = N \left( { \mu_0 I \over 2 a } \right) \Rightarrow N = 50 $
The distance at which the magnetic field on axis as compared to the mag. field at the centre of the coil carrying current I and radius R is 1/8 , would be




$ { B_1 \over B_2 } = \left( { R^2 \over X^2 + R^2 } \right) ^{3/2} $ $ 2^{-1} = \left( { R^2 \over X^2 + R^2 } \right) ^{1/2 } $ $ { 1 \over 8 } = \left( { R^2 \over X^2 + R^2 } \right) ^{3/2} $ $ { 1 \over 4} = \left ( { R^2 \over X^2 + R^2 } \right) $ $ 2^{-3} = \left( { R^2 \over X^2 + R^2} \right) ^{3/2} $ $ x = \sqrt 3 R $