With solenoid operated current will be due to inductance + coil resistance.

An important point from Andy's answer is that if significantly increased core saturation occurs due to the lower frequency then a substantial drop in impedance and increase in current may occur. My following comments apply IF the core does not saturate significantly more at 50 Hz than at 60 Hz - and it may.

Whether saturation increase is significant or not, and how to tell:

Solenoids tend to have some degree of airgap present due to imperfect mechanical closure at the 'striking face' and the existence of an air-gap between slider and stationary core. If desired these airgaps can be designed to be very small, but usually this is not a major requirement. The result is that solenoids may often be somewhat more tolerant to core saturation than transformers that have no or small well designed airgaps.
But, this is not certain and needs to be checked.
An "easy enough" check for core saturation increase is to measure operated current at 50 Hz and rated voltage and compare this with that specified for 60 Hz. If the increase is less than about 15 to 20% then the effects that I describe below are mainly involved. (You can calculate the likely current with no saturation change using my procedure described below.) If the increase is above or well above 20% then saturation is probably occurring significantly. In such cases a suitable series resistor that brings solenoid AC current closer to the desired range may be in order.

Current at 50 Hz will be greater than at 60 Hz due to lower impedance of inductor. This can be as much as 60/50 = 20% higher but will be somewhat less than this due to the coil resistance. Main losses will probably be I^2R losses in coil resistance and these will be up to 40% higher due to I^2 term.

Extra thermal losses may be OK depending on ambient temperature, enclosure and manufacturers ratings. Usually I'd expect it to be "probably OK but it may not be. look at resistance and rated operating current - which will show you what the effective resistance is to AC so how much the inductor is affecting things.

eg made up example.

Voperating = 220 VAC.
Ioperating = 100 mA.
Rcoil = 1000 Ohms.
So effective "AC resistance" = V/R = 220/0.1A = 2200 Ohms.
Coil impedance will be the vector sum of R + L impedances.
R_AC_effective = 2200 Ohm
R_R = 1000 Ohm
R_AC_coil = sqrt(Zl^2 + 1000^2) = 2200
so Zl = sqrt (2200^2 -1000^2) = 1960 Ohms.

So Zl 60Hz = 1960.
Zl 50 Hz = 50/60 x 1960 = 1633 Ohms.
Rcoil_AC_50Hz = sqrt(1633^2 + 1000^2) = 1914 Ohms AC.
I increas = 2200/1914 = 1.15x = 15%.
Power increase = 1.15^2 = 32%.
Prior power in Rl = R x I^2 = 1000 x 0.1A^2 = 10 Watts.
New power in Rl ~= 13 Watts.
Q: Does it matter?
A: Probably not, but, it may.

As above, if significant saturation is occurring then the current increase will be above that figure.
If current increase is more than 20% then saturation is definitely occurring (as 20% increase is the maximum with a purely inductive coil and no saturation).

A series resistor can be added to reduce current as desired - with possible affect on operation needing to be watched for with increaqsing vakues of resistance.

With solenoid operated current will be due to inductance + coil resistance.

An important point from Andy's answer is that if significantly increased core saturation occurs due to the lower frequency then a substantial drop in impedance and increase in current may occur. My following comments apply IF the core does not saturate significantly more at 50 Hz than at 60 Hz - and it may.

Whether saturation increase is significant or not, and how to tell:

Solenoids tend to have some degree of airgap present due to imperfect mechanical closure at the 'striking face' and the existence of an air-gap between slider and stationary core. If desired these airgaps can be designed to be very small, but usually this is not a major requirement. The result is that solenoids may often be somewhat more tolerant to core saturation than transformers that have no or small well designed airgaps.
But, this is not certain and needs to be checked.
An "easy enough" check for core saturation increase is to measure operated current at 50 Hz and rated voltage and compare this with that specified for 60 Hz. If the increase is less than about 15 to 20% then the effects that I describe below are mainly involved. (You can calculate the likely current with no saturation change using my procedure described below.) If the increase is above or well above 20% then saturation is probably occurring significantly. In such cases a suitable series resistor that brings solenoid AC current closer to the desired range may be in order.

Current at 50 Hz will be greater than at 60 Hz due to lower impedance of inductor. This can be as much as 60/50 = 20% higher but will be somewhat less than this due to the coil resistance. Main losses will probably be I^2R losses in coil resistance and these will be up to 40% higher due to I^2 term.

Extra thermal losses may be OK depending on ambient temperature, enclosure and manufacturers ratings. Usually I'd expect it to be "probably OK but it may not be. look at resistance and rated operating current - which will show you what the effective resistance is to AC so how much the inductor is affecting things.

eg made up example.

Voperating = 220 VAC.
Ioperating = 100 mA.
Rcoil = 1000 Ohms.
So effective "AC resistance" = V/R = 220/0.1A = 2200 Ohms.
Coil impedance will be the vector sum of R + L impedances.
R_AC_effective = 2200 Ohm
R_R = 1000 Ohm
R_AC_coil = sqrt(Zl^2 + 1000^2) = 2200
so Zl = sqrt (2200^2 -1000^2) = 1960 Ohms.

So Zl 60Hz = 1960.
Zl 50 Hz = 50/60 x 1960 = 1633 Ohms.
Rcoil_AC_50Hz = sqrt(1633^2 + 1000^2) = 1914 Ohms AC.
I increas = 2200/1914 = 1.15x = 15%.
Power increase = 1.15^2 = 32%.
Prior power in Rl = R x I^2 = 1000 x 0.1A^2 = 10 Watts.
New power in Rl ~= 13 Watts.
Q: Does it matter?
A: Probably not, but, it may.

As above, if significant saturation is occurring then the current increase will be above that figure.
If current increase is more than 20% then saturation is definitely occurring (as 20% increase is the maximum with a purely inductive coil and no saturation).

A series resistor can be added to reduce current as desired - with possible affect on operation needing to be watched for with increasing values of resistance.

I'll tidy this up "later". Duty calls. This provides enough guidance.

With solenoid operated current will be due to inductance + coil resistance. Current at 50 Hz will be greater than at 60 Hz due to lower impedance of inductor. This can be as much as 60/50 = 20% higher but will be somewhat less than this due to the coil resistance. Main losses will probably be I^2R losses in coil resistance and these will be up to 40% higher due to I^2 term.

Extra thermal losses may be OK depending on ambient temperature, enclosure and manufacturers ratings. Usually I'd expect it to be "probably OK but it may not be. look at resistance and rated operating current - which will show you what the effective resistance is to AC so how much the inductor is affecting things.
eg made up example.

Voperating = 220 VAC.
Ioperating = 100 mA.
Rcoil = 1000 Ohms.
So effective "AC resistance" = V/R = 220/0.1A = 2200 Ohms.
R + L impedances will vector sum ()Zcoil = sqrt(R^2+Zl^2) but usually close enough Zcoild = R + Zl (calculate actual values if desired)
So simple way Zl ~= 2200 - 1000 = 1200 Ohms.
So on 50 Hz Zl will be ~= 1200 x 50/60 = 1000 Ohms.
Overall Zcoil will then be R + Zl = 1000 + 1000 = 2000 Ohms.
So current will rise by 2200/2000 = +10%. Heating in Rcoil will rise 20%.

Let's try "proper" calculation above.

Reffective =220 Ohm.
Rl = 1000
sqrt(Zl^2 + 1000^2) = 2200
Zl = (sqrt 2200^2 -1000^2) = 1960 Ohms.
So above figure of Zl = 1200 Ohms was "too far off" :-(

So Zl 60Hz = 1960.
Zl 50 Hz = 50/60 x 1960 = 1633 Ohms.
Rcoil_AC = aqrt(1633^2 + 1000^2) = 1914 Ohms AC.
I increas = 2200/1914 = 1.15x = 15%.
Power increase = 32%.
Prior power in Rl = 1000 x 0.1A^2 = 10 Watts.
New power in Rl ~= 13 Watts.
Q: Does it matter?
A: Probably not, but, it may.

With solenoid operated current will be due to inductance + coil resistance.

An important point from Andy's answer is that if significantly increased core saturation occurs due to the lower frequency then a substantial drop in impedance and increase in current may occur. My following comments apply IF the core does not saturate significantly more at 50 Hz than at 60 Hz - and it may.

Whether saturation increase is significant or not, and how to tell:

Solenoids tend to have some degree of airgap present due to imperfect mechanical closure at the 'striking face' and the existence of an air-gap between slider and stationary core. If desired these airgaps can be designed to be very small, but usually this is not a major requirement. The result is that solenoids may often be somewhat more tolerant to core saturation than transformers that have no or small well designed airgaps.
But, this is not certain and needs to be checked.
An "easy enough" check for core saturation increase is to measure operated current at 50 Hz and rated voltage and compare this with that specified for 60 Hz. If the increase is less than about 15 to 20% then the effects that I describe below are mainly involved. (You can calculate the likely current with no saturation change using my procedure described below.) If the increase is above or well above 20% then saturation is probably occurring significantly. In such cases a suitable series resistor that brings solenoid AC current closer to the desired range may be in order. 

Current at 50 Hz will be greater than at 60 Hz due to lower impedance of inductor. This can be as much as 60/50 = 20% higher but will be somewhat less than this due to the coil resistance. Main losses will probably be I^2R losses in coil resistance and these will be up to 40% higher due to I^2 term.

Extra thermal losses may be OK depending on ambient temperature, enclosure and manufacturers ratings. Usually I'd expect it to be "probably OK but it may not be. look at resistance and rated operating current - which will show you what the effective resistance is to AC so how much the inductor is affecting things.

eg made up example.

Voperating = 220 VAC.
Ioperating = 100 mA.
Rcoil = 1000 Ohms.
So effective "AC resistance" = V/R = 220/0.1A = 2200 Ohms.
Coil impedance will be the vector sum of R + L impedances.
R_AC_effective = 2200 Ohm
R_R = 1000 Ohm
R_AC_coil = sqrt(Zl^2 + 1000^2) = 2200
so Zl = sqrt (2200^2 -1000^2) = 1960 Ohms.

So Zl 60Hz = 1960.
Zl 50 Hz = 50/60 x 1960 = 1633 Ohms.
Rcoil_AC_50Hz = sqrt(1633^2 + 1000^2) = 1914 Ohms AC.
I increas = 2200/1914 = 1.15x = 15%.
Power increase = 1.15^2 = 32%.
Prior power in Rl = R x I^2 = 1000 x 0.1A^2 = 10 Watts.
New power in Rl ~= 13 Watts.
Q: Does it matter?
A: Probably not, but, it may.

As above, if significant saturation is occurring then the current increase will be above that figure.
If current increase is more than 20% then saturation is definitely occurring (as 20% increase is the maximum with a purely inductive coil and no saturation).

A series resistor can be added to reduce current as desired - with possible affect on operation needing to be watched for with increaqsing vakues of resistance.