In the last problem, region to the right of CD is filled with a magnetic field of strength twice that in the region between AB and CD. This field exists up to a large distance and both fields are parallel in direction. For `d=((sqrt2-1)/(2))(mv)/(qB)` and `0=(pi)/(6),` find the displacement of the particle (measured from its entry point P) by the time it comes out of the magnetic fields. Find the total time spend by the particle in the two fields.
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A magnet is 10cm long and its pole strength is 120 CGS units (1 CGS unit of pole strength `= 0.1 A m)`. Find the magnitude of the magnetic field `B` a
The pole strength is `m=120 CGS units =12A-m` Magnetic length is `2l=10vm ` or `l=0.05 m`. Distance from the magnet is `d=20cm=0.2 m`.The field `B` at a point in end-on...
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A bar magnet has a pole strength of `3.6 A-m` and magnetic length `8 cm`.Find the magnetic at (a) a point on the axis at a distance of `6 cm` from the
Correct Answer - (a) `8.6xx10^(-4)T` ; (b) `7.7xx10^(-9)T`. `M=3.6xx8xx10^(2) A.m^(2)` (a) `B=mu_(0)/(4pi) (2Mr)/(r^(2)-a^(2))^(2)=8.6xx10^(-4) T`. , (b) `B=mu_(0)/(4pi)M/(r^(2)-a^(2))^(3//2)=7.7xx10^(-5) T`.
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Consider the situation of the previous problem. Find the average electric field energy stored in the capacitor and the average magnetic field energy s
Correct Answer - `25 mJ,5 mJ`
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Find (i) the last digit, (ii) the last two digits, and (iii) the last three digits of `17^(256)dot`
We have `17^(256) = (17^(2))^(128) = (289)^(128) = (290-1)^(128)` `:. 17^(256) = .^(128)C_(0)(290)^(128)-.^(128)C_(1)(290)^(127)+.^(128)C_(2)(290)^(126)-"....."` `- .^(128)C_(125)(290)^(3)+.^(128)C_(126)(290)^(2)-.^(128)C_(127)(290)+1` `=[.^(128)C_(0)(290)^(128) - .^(128)C_(1)(290)^(127)+.^(128)C_(2)(290)^(126)-"....."` `-.^(128)C_(125)(290)^(3)]+.^(128)C_(126)(290)^(2) -.^(128)C_(127)(290)+1` `=1000m+((128)(127))/(2)(290)^(2)-128xx290+1` `= 1000 m + (128)(127)(290)(145)-128xx290-1` `= 1000m+(128)(290)(127xx145-1)+1` `=1000m+(128)(290)(18414)+1` `=1000m+683527680+1`...
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Labour Problem, Marketing problem, Problem of transport, Domestic problem
Domestic problem
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In the last question, the electric field in vertical direction is switched off and a field of same strength`(E_0)` is switched on in horizontal direct
Correct Answer - `V_(xn)=(qE)/(m)sqrt((2h)/(g))[1+2e((1-e^(n-1))/(1-e))}` and `T=2e^(n)sqrt((2h)/(g))`
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calculate the force experinced by an electron when it enters a uniform magnetic field of strength 2 T with vleocity `0.2 xx 10^(8) m s^(-1)` at right
Correct Answer - `6.4 xx 10^(-6)` N; circular because magnetic force is always perpendicular to velocity
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Considering `theta=pi//2` as the zero-energy position of a current loop of magnetic moment m placed in a magnetic field of strength B,find the minimum
Correct Answer - `-mB` when `theta=0` , i.e. , coil phane is perpendicular to B yes, because `tau=0` in this position ; `mB(1-"coa"alpha)`
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Two bar magnets are in line. One has a magnetic length of `0.08m` and pole strength of 5Am and the other has a magnetic length of `0.04m` and pole str
Correct Answer - `7.4 xx 10^(-6)J`
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The region between ` x=o and x = L` is filled with uniform, steady magnetic field `B_(0) hat(k)`. A particle of mass `m` , positive charge `q` and vel
Correct Answer - [(a) `(mv_(0))/(2qB)` ; (b) `(pim)/(2qB)`]
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