Answer:
Magnitude of magnetic field is 1.29 x 10⁻⁴ T
Explanation:
Given :
Current flowing through the wire, I = 16.9 A
Length of wire. L = 0.69 m
Magnetic force experienced by the wire, F = 1.5 x 10⁻³ N
Consider B be the applied magnetic field.
The relation to determine the magnetic force experienced by current carrying wire is:
F = ILBsinθ
Here θ is the angle between magnetic field and current carrying wire.
According to the problem, the magnetic field and current carrying wire are perpendicular to each other, that means θ = 90⁰. So, the above equation becomes:
F = ILB

Substitute the suitable values in the above equation.

B = 1.29 x 10⁻⁴ T
Answer:A
Explanation:
In R-L circuit current is given by
![i=i_0\left [ 1-e^{\frac{-t}{L/R}}\right ]](https://tex.z-dn.net/?f=i%3Di_0%5Cleft%20%5B%201-e%5E%7B%5Cfrac%7B-t%7D%7BL%2FR%7D%7D%5Cright%20%5D)
where i=current at any time t

R=resistance
L=Inductance
at t=0
approaches to 1
therefore ![i=i_0\left [ 1-1\right ]](https://tex.z-dn.net/?f=i%3Di_0%5Cleft%20%5B%201-1%5Cright%20%5D)
i=0
when t approaches to
,
approaches to zero
thus 
thus we can say that initially circuit act as broken wire with zero current
and it increases exponentially with time and act as ordinary connecting wire
Answer:
Magnitude of the vector is
and the direction is 
Explanation:
Magnitude of first vector = 
Angle = 
Magnitude of second vector = 
Angle = 
x component of first vector

y component of first vector

x component of second vector

y component of first vector

Adding the magnitudes


Magnitude of the sum of the vectors would be

The direction would be

The magnitude of the vector is
and the direction is 