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51.

A `+6muC` point charge is moving at a constant velocity of `8xx10^6 ms^-1` in the +y direction, relative to a reference frame. At the instant when the point charge is at the origin of this reference frame, what is the magnetic field vector it produces at the following points? (a) `x=0.500m, y=0, z=0,` and (b) `x=0, y=-0.500m, z=0`.

Answer» For a charge with velocity
`vecv=(8.00xx10^-6m//s)hatj`,
the magnetic field produced at a position away from the particle
is `vecB=(mu_0)/(4pi) (qvecvxxhatr)/(r^2)`.
(a) `vecr=(+0.500m)hati implies hatvxxhatr=-hatk, r_0^2=1/4`
`vecB=-(mu_0)/(4pi) (qv)/(r_0^2) hatk`
`=-(mu_0)/(4pi) ((6.0xx10^-6C)(8.0xx10^6m//s))/((0.50m)^2)`
`=-(1.92xx10^-5T)hatk`
(b) `vecr=(-0.500m)hatj implies hatvxxhatr=0 implies vecB=0`
Discussion
52.

Two coaxil solenoids 1 and 2 of the same length are set so that one is inside the other. The number of turns per unit length are `n_(1)` and `n_(2)`. The current `i_(1)` and `i_(2)` are flowing in opposite directions. The magnetic field inside the inner coil is zero. This is possible whenA. `i_(1)ne i_(2)and n_(1) n_(2)`B. `i_(1)= i_(2)and n_(1) ne n_(2)`C. `i_(1)= i_(2)and n_(1)= n_(2)`D. `i_(1)= n_(2)and i_(1)n_(2)`

Answer» Correct Answer - C::D
Discussion
53.

In the figure, what is the magnetic field at the point `O`? A. `(mu_(0)I)/(4pir)`B. `(mu_(0)I)/(4pir)+(mu_(0)I)/(2pir)`C. `(mu_(0)I)/(4pi)+(mu_(0)I)/(4pir)`D. `(mu_(0)I)/(4pi)-(mu_(0)I)/(4pir)`

Answer» Correct Answer - C
Discussion
54.

A conducting ring of radius r having charge q is rotating with angular velocity `omega` about its axes. Find the magnetic field at the centre of the ring.

Answer» Current in the ring `I=(omegaq)/(2pi)`
Magnetic field `B=(mu_0I)/(2r)=(mu_0 omegaq)/((2pi)xx2r)`
` implies B=(mu_0 omegaq)/(4pir)`
Discussion
55.

A charge q coulomb moves in a circle at n revolution per second and the radius of the circle is r metre. Then magnetic feild at the centre of the circle isA. `(2piq)/(nr)xx10^-7NA^-1m^-1`B. `(2piq)/rxx10^-7NA^-1m^-1`C. `(2pinq)/rxx10^-7NA^-1m^-1`D. `(2piq)/rxx10^-7NA^-1m^-1`

Answer» Correct Answer - C
(c) `B=(mu_0I)/(2r) or B=(mu_0q)/(2rT)`
`B=(4pixx10^-7q)/(2r)n`
`B=(2pinq)/2xx10^-7NA^_1m^-1`
Discussion
56.

A particle is moving wirh velocity `vecv=hati+3hatj` and it produces an electric field at a point given by `vecE=2hatk`. It will produce magnetic field at that point equal to (all quantities are in SI units)A. `(6hati-2hatj)mu_0epsilon_0`B. `(6hati+2hatj)mu_0epsilon_0`C. zeroD. cannot be determine

Answer» Correct Answer - A
(a) `vecE=1/(4piepsilon_0) q/(r^2) hatr=2hatk,hatr=hatk`
`vecB=(mu_0)/(4pi) (qvecvxxhatr)/(r^2)=mu_0epsilon_0[q/(4piepsilonr^2)](hati+3hatj)xxhatk`
`=mu_0epsilon_02[-hatj+3hati]=(6hati-2hatj)mu_0epsilo_0`
Discussion
57.

A point chanrge of magnitude `q=4.5nC` is moving with speed `v=3.6xx10^7ms^-1` parallel to the x-axis along the line `y=3m`. Find the magnetic field at the origin produced by this charge when the charge is at the point `x=-4 m`, `y=3m`, as shown in Fig.

Answer» The magnetic field is given by
`vecB=(mu_0)/(4pi) (qvecvxxhatr)/(r^2)`, with `vecv=vhati`
`vecr=(4hati-3hatj)m implies vecr=sqrt(4^2+3^2)m=5m`
Unit vector in the direction of `hatr=(vecr)/(vecr)=(4hati-3hatj)/5=0.8hati-0.6hatj`
`vecvxxhatr=(vhati)xx(0.8hati-0.6hatj)=-0.6vhatk`
`vecB=(mu_0)/(4pi) (qvecvxxhatr)/(r^2)=(mu_0)/(4pi) (q(-0.6vhatk))/(r^2)`
`=-(10^-7)((4.5xx10^-19)(0.6)(3.6xx10^7))/(5^2)hatk`
`=-3.89xx10^-10 Thatk`
Discussion