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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 $
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.
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.