A uniform magnetic field exists in a region given by →B=3^i+4^j+5^k. A rod of length 5m along y-axis moves with a constant speed of 1m s−1 along x axis. Then the induced emf in the rod will be
A uniform magnetic field exists in a region given by →B=3^i+4^j+5^k. A rod of length 5m along y-axis moves with a constant speed of 1m s−1 along x axis. Then the induced emf in the rod will be
The correct option is B 25V
Given,B=3+4+5k
V=1m/s(As the rod moves in X−Direction)
Length of rod=5m
Since,e=−dϕ/dt and ϕ=B.A
So,e=−d(B.A)/dt
e=d(B.A)/dt(Negative sign is neglected as it only represents the direction)
e=B.d(A)/dt(Magnetic Field is constant)
=B.d(l×b)/dt(Where l=Length and b=Breadth)
=lB.d(b)/dt−−−−(1)
As db/dt is nothing but the velocity with which the rod is moving.
So,db/dt=V
Putting the above value of db/dt in equation (1)
We have, e=lB.V
But, in the above equation it is assumed that the direction of B, L and V are mutually perpendicular.
So, Only that component of B that is in mutual perpendicular direction of L and V will be taken into account.
So,e=lB.V=5×5×1
e=25V
Given,B=3+4+5k
V=1m/s(As the rod moves in X−Direction)
Length of rod=5m
Since,e=−dϕ/dt and ϕ=B.A
So,e=−d(B.A)/dt
e=d(B.A)/dt(Negative sign is neglected as it only represents the direction)
e=B.d(A)/dt(Magnetic Field is constant)
=B.d(l×b)/dt(Where l=Length and b=Breadth)
=lB.d(b)/dt−−−−(1)
As db/dt is nothing but the velocity with which the rod is moving.
So,db/dt=V
Putting the above value of db/dt in equation (1)
We have, e=lB.V
But, in the above equation it is assumed that the direction of B, L and V are mutually perpendicular.
So, Only that component of B that is in mutual perpendicular direction of L and V will be taken into account.
So,e=lB.V=5×5×1
